Advanced Calculus

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'Advanced Calculus 0

Fifth Edition I

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[ # ] Wilfred Kaplan

8

Preface to the Fifth Edition

As I fecall, it was in 1948 that Mark Morkovin, a colleague in engineering, approached me to suggest that I write a text for engineering students needing to proceed beyond elementary calculus to handle the new applications of mathematics. World War I1 had indeed created many new demands for mathematical skills in a variety of fields. Mark was persuasive and I prepared a book of 265 pages, which appeared in lithoprinted form, and it was used as the text for a new course for third-year students. The typesetting was done using a "varityper," a new typewriter that had keys for mathematical symbols. In the summer of 1949 I left Ann Arbor for a sabbatical year abroad, and we rented our home to a friend and colleague Eric Reissner, who had a visiting appointment at the University of Michigan. Eric was an adviser to a new publisher, Addison-Wesley, and learned about my lithoprinted book when he was asked to teach a course using it. He wrote to me, asking that I consider having it published by Addison-Wesley. Thus began the course of this book. For the first edition the typesetting was carried out with lead type and I was invited to watch the process. It was impressive to see how the type representing the square root of a function was created by physically cutting away at an appropriate type showing the square root sign and squeezing type for the function into it. How the skilled person carrying this out would have marveled at the computer methods for printing such symbols!

This edition differs from the previous one in that the chapter on ordinary differential equations included in the third edition but omitted in the fourth edition has been restored as Chapter 9. Thus the present book includes all the material present in the previous editions, with the exception of the introductory review chapter of the first edition. A number of minor changes have been made throughout, especially some updating of the references. The purpose of including all the topics is to make the book more useful for reference. Thus it can serve both as text for one or more courses and as a source of information after the courses have been completed.

ABOUTTHE BOOK The background assumed is that usually obtained in the freshman-sophomore calculus sequence. Linear algebra is not assumed to be known but is developed in the first chapter. Subjects discussed include all the topics usually found in texts on advanced calculus. However. there is more than the usual emphasis on applications and on physical motivation. Vectors are introduced at the outset and serve at many points to indicate geometrical and physical significance of mathematical relations. Numerical methods are touched upon at various points, both because of their practical value and because of the insights they give into the theory. A sound level of rigor is maintained throughout. Definitions are clearly labeled as such and all important results are formulated as theorems. A few of the finer points of real variable theory are treated at the ends of Chapters 2, 4, and 6. A large number of problems (with answers) are distributed throughout the text. These include simple exercises as well as complex ones planned to stimulate critical reading. Some -points of the theory are relegated to the problems. with hints given where appropriate. Generous references to the literature are given, and each chapter concludes with a list of books for supplementary reading. Starred sections are less essential in a first course.

Chapter 1 opens with a review of "ectors in space. determinants, and linear equations, and then develops matrix algebra, including Gaussian elimination, and n-dimensional geometry, with stress on linear mappin.gs.The second chapter takes up partial derivatives and develops them with the aid of vectors (gradient, for example) and matrices; partial derivatives are applied to geometry and to maximum-minimum problems. The third chapter introduces divergence and curl and the basic identities; orthogonal coordinates are treated concisely; final sections provide an introduction to tensors in n-dimensional space. The fourth chapter, on integration, reviews definite and indefinite integrals, using numerical methods to show how the latter can be constructed; multiple integrals are treated carefully, with emphasis on the rule for change of variables; Leibnitz's Rule for differentiating under the integral sign is proved. Improper integrals are also covered; the discussion of these is completed at the end of Chapter 6, where they are

related to infinite series. Chapter 5 is devoted to line and surface integrals. Although the notions are first presented without vectors, it very soon becomes clear how natural the vector approach is for this subject. Line integrals are used to provide an exceptionally complete treatment of transformation of variables in a double integral. Many physical applications. including potential theory, are given. Chapter 6 studies infinite series without assumption of previous knowledge. The notions of upper and lower limits are introduced and used sparingly as a simplifying device; with their aid, the theory is given in almost coinplete form. The usual tests are given: in particular, the root test. With its aid, the treatment of power series is greatly simplified. Uniform convergence is presented with great care and applied to power series. Final sections point out the parallel with improper integrals; in particular, power series are shown to correspond to the Laplace transform. Chapter 7 is a complete treatment of Fourier series at an elementary level. The first sections give a simple introduction with many examples; the approach is gradually deepened and a convergence theorem is proved. Orthogonal functions are then studied, with the aid of inner product, norm, and vector procedures. A general theorem on complete systems enables one to deduce completeness of the trigonometric system and Legendre polynomials as a corollary. Closing sections cover Bessel functions, Fourier integrals, and generalized functions. Chapter 8 develops the theory of analytic functions with emphasis on power series, Laurent series and residues, and their applications. It also provides a full treatment of conformal mapping, with many examples and physical applications and extensive discussion of the Dirichlet problem. Chapter 9 assumes some background in ordinary differential equations. Linear systems are treated with the aid of matrices and applied to vibration problems. Power series methods are treated concisely. A unified procedure is presented to establish existence and uniqueness for general systems and linear systems. The final chapter, on partial differential equations, lays great stress on the relationship between the problem of forced vibrations of a spring (or a system of springs) and the partial differential equation

By pursuing this idea vigorously the discussion uncovers the physical meaning of the partial differential equation and makes the mathematical tools used become natural. Numerical methods are also motivated on a physical basis. Throughout, a number of references are made to the text Culci~lusand Linear Algebru by Wilfred Kaplan and Donald J. Lewis (2 vols., New York, John Wiley & Sons, 1970-197 l), cited simply as CLA.

SUGGESTIONS ON THE USEOF THISBOOKAS THE TEXTFOR A COURSE The chapters are independent of each other in the sense that each can be started with a knowledge of only the simplest notions of the previous ones. The later portions of the chapter may depend on some of the later portions of eailier ones. It is thus possible to construct a course using just the earlier portions of several chapters. The following is an illustration of a plan for a one-semcster course, meeting four hours

a week: 1.1 to 1.9, 1:14, 1.16, 2.1 to 2.10,2.12 td2.18, 3.1 to 3.6,4.1 to 4.9, 5.1 to 5.13, 6.1 to 6.7, 6.11 to 6.19. If it is d e & d that one topic be stressed, then the corresponding chapters can be taken up in full detail.F& example, Chapters 1,3, and 5 together provide a very substantial training in vector analysis; Chapters 7 and 10 together contain sufficient material for a one-semester course in partial differential equations; Chapter 8 provides sufficient text for a one-semester course in complex variables. I express my appreciation to the many colleagues who gave advice and encouragement in the preparation of this book. Professors R. C. F. Bartels, F. E. Hohn, and J. Lehner deserve special thanks and recognition for their thorough criticisms of the first manuscript: a number of improvements were made on the basis of their suggestions. Others whose counsel has been of value are Professors R.V. Churchill, C. L. Dolph, G. E. Hay, M. Morkovin, G. Piranian, G. Y. Rainich, L. L. Rauch, M. 0. Reade, E. Rothe, H. Samelson, R. Buchi, A. J. Lohwater, W. Johnson, and Dr. G. BCguin. For the preparation of the third edition, valuable advice was provided by Professors James R. Arnold, Jr., Douglas Cameron, Ronald Guenther, Joseph Horowitz, .and David 0. Lomen. Similar help was given by Professors William M. Boothby, Harold Parks, B. K. Sachveva, and M. Z. Nashed for the fourth edition and by Professors D. Burkett, S. Deckelman. L. Geisler, H. Greenwald, R. Lax, B. Shabell and M. Smith for the present edition. To Addison-Wesley publishers 1 take this occasion to express my appreciation for their unfailing support over many decades. Warren Blaisdell first represented them, and his energy and zeal did much to get the project under way. Over the years many others carried on the high standards he had set. I mention David Geggis, Stephen Quigley, and Laurie Rosatone as ones whose fine cooperation was typical of that provided by the company. To my wife I express my deeply felt appreciation for her aid and counsel in every phase of the arduous task and especially for maintaining her supportive role for this edition, even when conditions have been less than ideal. Wilfred Kaplan Ann Arbor, Michigan

Contents

Vectors and Matrices 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 *1.11 *1.12 *1.13 1.14 *1.15 1.16

Introduction 1 Vectors in Space 1 Linear Independence rn Lines and Planes 6 Determinants 9 Simultaneous Linear Equations 13 Matrices 18 Addition of Matrices Scalar Times Matrix 19 Multiplication of Matrices 21 Inverse of a Square Matrix 26 Gaussian Elimination 32 Eigenvalues of a Square Matrix 35 The Transpose 39 Orthogonal Matrices 41 Analytic Geometry and Vectors in n-Dimensional Space Axioms for Vn 51 Linear Mappings 55

46

*1.17 Subspaces rn Rank of a Matrix * l . l 8 Other Vector Spaces 67

62

Differential Calculus of Functions of Several Variables 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 *2.11 2.12 2.13 2.14 2.15 2.16 2.17 2.18 2.19 *2.20

73

Functions of Several Variables 73 Domains and Regions 74 Functional Notation rn Level Curves and Level Surfaces 76 Limits and Continuity 78 Partial Derivatives 83 Total Differential Fundamental Lemma 86 Differential of Functions of n Variables The Jacobian Matrix 90 Derivatives and Differentials of Composite Functions 96 The General Chain Rule 101 Implicit Functions 105 Proof of a Case of the Implicit Function Theorem 112 Inverse Functions Curvilinear Coordinates 118 Geometrical Applications 122 The Directional Derivative 131 Partial Derivatives of Higher Order 135 Higher Derivatives of Composite Functions 138 The Laplacian in Polar, Cylindrical, and Spherical Coordinates 140 Higher Derivatives of Implicit Functions 142 Maxima and Minima of Functions of Several Variables 145 Extrema for Functions with Side Conditions Lagrange Multipliers 154 155 *2.21 Maxima and Minima of Quadratic Forms on the Unit Sphere *2.22 Functional Dependence 161 *2.23 Real Variable Theory rn Theorem on Maximum and Minimum 167

Vector Differential Calculus 3.1 3.2 3.3 3.4 3.5

Introduction 175 Vector Fields and Scalar Fields 176 The Gradient Field 179 The Divergence of a Vector Field 181 The Curl of a Vector Field 182

3.6 *3.7 *3.8 *3.9 *3.10 *3.11

Combined Operations 183 Curvilinear Coordinates in Space rn Orthogonal Coordinates Vector Operations in Orthogonal Curvilinear Coordinates Tensors 197 Tensors on a Surface or Hypersurface 208 Alternating Tensors rn Exterior Product 209

187 190

Integral Calculus of Functions of Several Variables 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 *4.10 *4.11

The Definite Integral 2 15 Numerical Evaluation of Indefinite Integrals rn Elliptic Integrals Double Integrals 225 Triple Integrals and Multiple Integrals in General 232 Integrals of Vector Functions 233 Change of Variables in Integrals 236 Arc Length and Surface Area 242 Improper Multiple Integrals 249 Integrals Depending on a Parameter Leibnitz's Rule 253 Uniform Continuity rn Existence of the Riemann Integral 258 Theory of Double Integrals 261

Vector Integral Calculus Two-Dimensional Theory 5.1 5.2 5.3 5.4 5.5 5.6 5.7

Introduction 267 Line Integrals in the Plane 270 Integrals with Respect to Arc Length Basic Properties of Line Integrals 276 Line Integrals as Integrals of Vectors 280 Green's Theorem 282 Independence of Path rn Simply Connected Domains 287 Extension of Results to Multiply Connected Domains 297

Three-Dimensional Theory and Applications 5.8 Line Integrals in Space 303 5.9 Surfaces in Space rn Orientability 5.10 Surface Integrals 308

305

215 22 1

The Divergence Theorem 3 14 Stokes's Theorem 32 1 Integrals Independent of Path Irrotational and Solenoidal Fields, Change of Variables in a Multiple Integral 33 1 Physical Applications 339 350 Potential Theory in the Plane Green's Third Identity 358 Potential Theory in Space 36 1 Differential Forms 364 Change of Variables in an m-Form and General Stokes's Theorem Tensor Aspects of Differential Forms 370 Tensors and Differential Forms without Coordinates 37 1

Infinite Series 6.1 6.2 6.3 6.4 6.5 6.6 6.7 *6.8 *6.9 6.10 6.1 1 6.12 6.13 6.14 6.15 6.16 6.17 6.18 *6.19 *6.20 *6.21 *6.22 *6.23 .8.

Introduction 375 Infinite Sequences 376 Upper and Lower Limits 379 Further Properties of Sequences 38 1 Infinite Series 383 Tests for Convergence and Divergence 385 Examples of Applications of Tests for Convergence and Divergence 392 Extended Ratio Test and Root Test 397 Computation with Series Estimate of Error 399 Operations on Series 405 Sequences and Series of Functions 4 10 Uniform Convergence 41 I Weierstrass M-Test for Uniform Convergence 416 Properties of Uniformly Convergent Series and Sequences 4 18 Power Series 422 Taylor and Maclaurin Series 428 Taylor's Formula with Remainder 430 Further Operations on Power Series 433 Sequences and Series of Complex Numbers 438 Sequences and Series of Functions of Several Variables 442 Taylor's Formula for Functions of Several Variables 445 Improper Integrals Versus Infinite Series 447 Improper Integrals Depending on a Parameter Uniform Convergence 453

325

368

*6.24 Principal Value of Improper Integrals 455 *6.25 Laplace Transformation rn r-Function and B-Function *6.26 Convergence of Improper Multiple Integrals 462

457

Fourier Series and Orthogonal Functions 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 *7.11 *7.12 *7.13 *7.14 *7.15 *7.16 *7.17 *7.18 *7.19 *7.20

Trigonometric Series 467 Fourier Series 469 Convergence of Fourier Series 47 1 Examples rn Minimizing of Square Error 473 Generalizations rn Fourier Cosine Series Fourier Sine Series 479 Remarks on Applications of Fourier Series 485 Uniqueness Theorem 486 Proof of Fundamental Theorem for Continuous, Periodic, and Piecewise Very Smooth Functions 489 Proof of Fundamental Theorem 490 Orthogonal Functions 495 Fourier Series of Orthogonal Functions rn Completeness 499 Sufficient Conditions for Completeness 502 Integration and Differentiation of Fourier Series 504 Fourier-Legendre Series 508 Fourier-Bessel Series 512 Orthogonal Systems of Functions of Several Variables 517 Complex Form of Fourier Series 5 18 Fourier Integral 5 19 The Laplace Transform as a Special Case of the Fourier Transform 521 Generalized Functions 523

Functions of a Complex Variable 8.1 8.2 8.3 8.4 8.5 8.6 8.7

Complex Functions 53 1 Complex-Valued Functions of a Real Variable 532 Complex-Valued Functions of a Complex Variable rn Limits and Continuity 537 Derivatives and Differentials 539 Integrals 541 Analytic Functions B-Cauchy-RiemannEquations 544 The Functions log z , a', z a , sin-' z , cos-' z 549

Integrals of Analytic Functions w Cauchy Integral Theorem 553 Cauchy's Integral Formula 557 Power Series as Analytic Functions 559 Power Series Expansion of General Analytic Function 562 Power Series in Positive and Negative Powers Laurent Expansion 566 Isolated Singularities of an Analytic Function w Zeros and Poles 569 The Complex Number oo 572 Residues 575 Residue at Infinity 579 Logarithmic Residues Argument Principle 582 Partial Fraction Expansion of Rational Functions 584 Application of Residues to Evaluation of Real Integrals 587 Definition of Conformal Mapping 59 1 Examples of Conformal Mapping 594 Applications of Conformal Mapping w The Dirichlet Problem 603 Dirichlet Problem for the Half-Plane 604 Conformal Mapping in Hydrodynamics 6 12 6 14 Applications of Conformal Mapping in the Theory of Elasticity Further Applications of Conformal Mapping 616 General Formulas for One-to-one Mapping Schwarz-Christoffel Transformation 6 17

Ordinary Differential Equations 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9

Differential Equations 625 Solutions 626 The Basic Problems Existence Theorem 627 Linear Differential Equations 629 Systems of Differential Equations w Linear Systems 636 Linear Systems with Constant Coefficients 640 A Class of Vibration Problems 644 Solution of Differential Equations by Means of Taylor Series The Existence and Uniqueness Theorem 65 1

Partial Differential Equations 10.1 10.2

Introduction 659 Review of Equation for Forced Vibrations of a Spring

646

659 66 1

Case of Two Particles 662 Case of N Particles 668 Continuous Medium Fundamental Partial Differential Equation 674 Classification of Partial Differential Equations w Basic Problems 676 The Wave Equation in One Dimension Harmonic Motion 678 Properties of Solutions of the Wave Equation 681 685 The One-Dimensional Heat Equation w Exponential Decay Properties of Solutions of the Heat Equation 687 Equilibrium and Approach to Equilibrium 688 Forced Motion 690 695 Equations with Variable Coefficients w Sturm-Liouville Problems Equations in Two and Three Dimensions Separation of Variables 698 Unbounded Regions Continuous Spectrum 700 Numerical Methods 703 Variational Methods 705 Partial Differential Equations and Integral Equations 707

Answers to Problems Index

713

Vectors and Matrices

Our main goal in this book is to develop higher-level aspects of the calculus. The calculus deals with functions of one or more variables. The simplest such functions are the linear ones: for example, y = 2x 5 and z = 4x 7y 1. Normally, one is forced to deal with functions that are not linear. A central idea of the differential calculus is the approximation of a nonlinear function by a linear one. Geometrically, one is approximating a curve or surface or similar object by a tangent line or plane or similar linear object built of straight lines. Through this approximation, questions of the calculus are reduced to ones of the algebra associated with lines and planeslinear algebra. This first chapter develops linear algebra with these goals in mind. The next four sections of the chapter review vectors in space, determinants, and simultaneous linear equations. The following sections then develop the theory of matrices and some related geometry. A final section shows how the concept of vector can be generalized to the objects of an arbitrary "vector space."

+

+ +

We assume that mutually perpendicular x , y, and z axes are chosen as in Fig. 1.1, and we assume a common unit of distance along these axes. Then every point P in space has coordinates (x , y , z ) with respect to these axes, as in Fig. 1.1. The origin 0 has coordinates (0, 0, 0).

2

Advanced Calculus. Fifth Edition

Figure 1.1

Coordinates in space.

A vector v in space has a magnitude (length) and direction but no fixed location. We can thus represent v by any one of many directed line segments in space, all having the same length and direction (Fig. 1.1). In particular, we can represent v by the directed line segment from 0 to a point P , provided that the direction from 0 to P is that of v and that the distance from 0 to P equals the length of v, as suggested in Fig. 1.1. We write simply

The figure also shows the components u,, v,, v, of v along the axes. When (1.1) holds, we have

We assume the reader's familiarity with addition of vectors and multiplication of vectors by numbers (scalars). With the aid of these operations a general vector v can be represented as follows:

Here i, j, k are unit vectors (vectors of length 1) having the directions of the coordinate axes, as in Fig. 1.2. By the Pythagorean theorem, v then has magnitude, denoted by I vl, given by the equation

4

In particular, for v = OP the distance of P : (.r, y , z ) from 0 is

131 = J-.

(1.5)

Chapter 1 Vectors and Matrices

Figure 1.2

Vector v in terms of i, j, k.

Figure 1.3

Definition of dot product.

More generally, for v = PI6 ,where PI is ( X I yl, , 2 , ) and P2 is (x2,y2, z2),one has * * v = OP2 - OP1 = (x2 - xl)i . . . and the distance between PI and P2 is

+

The vector v can have 0 length, in which case v = with 0 .We then write

3 only when P coincides

and call v the zero vector. The vector v is completely specified by its components v,, v,, v,. It is often convenient to write

instead of Eq. (1.3). Thus we think of a vector in space as an ordered triple of numbers. Later we shall consider such triples as matrices (row vectors or column vectors). The dot product (or inner product) of two vectors v, w in space is the number

, chosen between 0 and n inclusive (see Fig. 1.3). When v or w where 0 = ~ ( vw), is 0, the angle 6 is indeterminate, and v . w is taken to be 0. We also have v . w = 0 when v, w are orthogonal (perpendicular) vectors, v Iw. We agree to say that the 0 vector is orthogonal to all vectors (and parallel to all vectors). With this conventi

3

4

Advanced Calculus, Fifth Edition

Figure 1.4

Component.

we can state: v . w = 0 precisely when

v l w.

The dot product satisfies some algebraic rules: u.v=v.u.

u.(v+w)=u~v+u.W,

u . (cv) = (cu) . v = C(U. v),

U

. U = 1u12 .

(1.11)

Here c is a scalar. In Eq. (1.9) the quantity Ivl cos 0 is interpreted as the component of v in the direction of w (see Fig. 1.4):

This can be positive, negative, or 0. The angles a, B, and y , between v (assumed to be nonzero) and the vectors i, j, and k , respectively, are called direction angles of v; the corresponding cosines cos a , cos B, cos y are the direction cosines of v. By Eqs. (1.9) and (1.12) and Fig. 1.2,

Accordingly,

Thus the vector (I / /vl)v has components cos a, cos B, cos y ; we observe that this vector is a unit vector, since its length is (1 //v()lvl = 1. Since i . i = 1, i . j = 0, etc., we can compute the dot product of

u = u,i

+ uyj+ uzk

and

v = v,i

+ v,j + v,k

as follows: u . v = (u,i

+ uyj + u z k ) .(v,i + vyj + v,k)

= u , v , i ~ i + u , v , i ~ j +....

Here we use the rules ( 1.1 1). We conclude:

Chapter 1 Vectors and Matrices

Figure 1.5

Vector product.

The vector product or cross product u x v of two vectors u, v is defined with reference to a chosen orientation of space. This is usually specified by a right-handed xyz-coordinate system in space. An ordered triple of vectors is then called apositive triple if the vectors can be moved continuously to attain the respective directions of i, j, k eventually without making one of the vectors lie in a plane parallel to the other two; a practical test for this is by aligning the vectors with the thumb, index finger, and middle finger of the right hand. The triple is called negative if the test can be satisfied by using j, i, k instead of i, j, k. If one of the vectors is 0 or all three vectors are coplanar (can be represented in one plane), the definition is not applicable. Now we define u x v = w, where

lwl = lul Ivl sin0.

0 = <(u, v),

(1.16)

u, v, w form a positive triple. This is illustrated in Fig. 1.5. The definition breaks down when u or v is 0 or when 8 = 0 or ~ ( uv ,collinear). In these cases we write u x v = 0. We can say simply: u x v = 0 precisely when ullv.

(1.17)

From Eq. ( I . 16) we observe that

= area of parallelogram of sides u, v,

as illustrated in Fig. 1.5. The vector product satisfies algebraic rules:

u x (rv) = (cu) X v = c(u x v),

The last two rules are described as the identities for vector triple products. Since i x i = 0, i x j = k,i x k I -j, and so on, we can calculate u x v as

5

6

Advanced Calculus. Fifth Edition and conclude:

This can also be written as a determinant (Section 1.4):

Here we expand by minors of the first row. From the rules ( I . 19) we see that, in general, u x v # v x u and u x (v x w) # (u X v) X W. For further discussion of vectors, see Chapter 11 of CLA.'

1.3 LINEARINDEPENDENCE

8

LINESAND PLANES

Two vectors u, v in space are said to be linearly independent if they cannot be represented by directed line segments on the same line. Otherwise, they are said to be linearly dependent or collinear (see Fig. 1.6). When u or v is 0, the vectors are considered to be linearly dependent. We thus see that u, v are linearly dependent precisely when u x v = 0. Three vectors u, v, w in space are said to be linearly independent when they cannot be represented by directed line segments in the same plane. Otherwise, they are said to be linearly dependent or coplanar (see Fig. 1.7). We can include both these cases in a general definition: Vectors u l , . . . , uk in space are linearly independent if the only scalars c l . . . . , ck such that

a r e e l = 0 , c 2 = 0, . . . , ck = 0. For k = 2, ul and ul are thus linearly dependent if c l u l scalars cl. c2 that are not both 0. If, say, c2 # 0, then

+ c2uz= 0 for some

Thus u-, is a scalar times ul and is collinear with u l (if e l = 0 or ul = 0, then u2 would be 0). Conversely, if ul , u2 are collinear, then u2 - kul = 0 or ul - ku2 = 0 for some scalar k. Thus the new definition agrees with the old one. Similarly, fork = 3, u l ,u2,u, are linearly dependent if c l u l c2u2 c3u3= 0 for some scalars e l ,c2, c3 that are not all 0. If, for example, c3 # 0, then

+

+

Thus u3 is a linear iorrzbination oCul and u2,so the three must be coplanar (Fig. 1.8). h he work Calculus und Lineur Algrbru by the author and Donald J . Lewis, 2 vols. (New York: John Wiley and Sons, Inc., 1970-1971), will be referred to throughout as CLA.

Chapter 1

Vectors and Matrices

Figure 1.6 (a) Linearly independent vectors u, v. (b) Linearly dependent vectors u, v.

Figure 1.7 (a) Linearly independent vectors u, v. w. (b) Linearly dependent vectors u, v, w.

Figure 1.8

The vector uj as a linear combination of u , , u2,

Conversely, if the three vectors are coplanar, then it can be verified that one must equal a linear combination of the other two, say, u3 = klul k2u2and then klul kzu:! - lu3 = 0, SO the vectors are linearly dependent by the new definition. Again the two definitions agree. What about four vectors in space? Here the answer is simple: They must be linearly dependent. Let the vectors be ul, u:!, u3, u4. There are then two possibilities: (a) u l , u2,u3 are linearly dependent, and (b) u , , u2,us are linearly independent. In case (a),

+

+

for some scalars c l , c?.c not all 0. But then

with not all of cl, ~ 2 ~3, equal to 0. Thus u l . u2,~ 3 u4. are linearly dependent. In case (b), ul, u2, U? are not coplanar and hence can be represented by the directed edges of a parallelepiped in space, as in Fig. 1.9. From this it follows that u4 can

7

Advanced Calculus. Fifth Edition

Figure 1.9

Expression of u4 as a linear combination of u,,u2,u,.

+

+

be represented as c l u l ~ 2 ~~ 23 for~ appropriate 3 e l , ~ 2 c3, , as in the figure; this is analogous to the representation of v in terms of i, j, k in Eq. (1.3) and Fig. 1.2. Now

so that again u l , . . . , ~q are linearly dependent. Accordingly, there cannot be four linearly independent vectors in space. By similar reasoning we see that for every k greater than 3 there is no set of k linearly independent vectors in space. However, for k 5 3 there are k linearly independent vectors in space. For example, i, j is such a set of two vectors, and i, j, k is such a set of three vectors. (We can also consider i by itself-or any nonzero vector-as a set of one linearly independent vector.) Every triple u l , uz, u3 of linearly independent vectors in space serves as a basis for vectors in space; that is, every vector in space can be expressed uniquely as a linear combination c l u l c2u2 c3u3,as in Fig. 1.9. v,j v,k is the We call i, j, k the standard basis. The equation v = v,i representation of v in terms of the standard basis. We observe that one could specialize the discussion of linear independence to two-dimensional space-that is, the xy-plane. Here there are pairs of linearly independent vectors, and each such pair forms a basis; i, j is the standard basis. Every set of more than two vectors in the plane is linearly dependent.

+

+

+

+

+ +

Planes in space. If PI: (xl, y l , z l ) is a point of a plane and n = Ai Bj Ck is a nonzero normal vector (perpendicular to the plane), then P : (x, y, z) is in the plane precisely when

(see Fig. 1.10). Equation (1.24) can be written as a linear equation

Chapter 1 Vectors and Matrices

Figure 1.10

Plane.

Figure 1.11

Line, distance s as parameter.

and every linear equation (1.25) (A, B, C not all 0) represents a plane, with n = Ai Bj + C k as normal vector.

+

Lines in space. If PI: (xl, yl , z,) is a point of a line and v = ai+ bj + c k is a nonzero vector along the line (that is, representable by a directed line segment joining two points of the line), then P : (x, y, z) is on the line precisely when

that is, when v and times v:

$ are linearly dependent. Since v # 0, $ must be a scalar

where t can be any number. From Eq. (1.27) we obtain parametric equations of the line: x =XI +at,

y = yl

+ bt,

z =zl

+ ct,

--a
oo.

i

(1.28)

If v happens to be a unit vector, then = Itl, so that t can be regarded as a distance coordinate along the line. In this case we usually replace t by s , so that x

=XI

+as,

y = yl

+ bs,

z = zl

+ cs,

as in Fig. 1.11.

1.4 DETERMINANTS For second-order determinants, one has the formula

-a< s < oo, (1.29)

9

10

Advanced Calculus. Fifth Edition Then higher-order determinants are reduced to those of lower order. For example,

From these formulas, one sees that a determinant of order n is a sum of terms, each of which is f1 times a product of n factors, one each from the n columns of the array and one each from the n rows of the array. Thus from (1.3 1) and (1.30), one obtains the six terms

We now state six rules for determinants:

I. Rows and columns can be interchanged. For example,

: ;1 ; ;1=!

6 1 c~

a2

:;I.

a3

Hence in every rule the words row and column can be interchanged. 11. Interchanging two rows (or columns) multiplies the determinant by -1. For example,

Ill. A factor of any row (or column) can be placed before the determinant. For example,

IV lf two rows (or columns) are proportional, the determinant equals 0. For example,

1

1

kal kbl kcl a bl c~ = 0. a2 b2 C2 V Determinants differing in only one row (or column) can be added by adding corresponding elements in that row and leaving the other elements unchanged. For example,

Chapter 1 Vectors and Matrices VI. The value of a determinant is unchanged if the elements of one row are multipled by the same quantity k and added to the corresponding elements of another row. For example,

, I : =I bl

cl

a,

+ ka2

:

bl

+ kb2 b3 b2

CI

+ kc2 c3 c2

1.

By a suitable choice of k , one can use this rule to introduce zeros; by repetition of the process, one can reduce all elements but one in a chosen row to 0. This procedure is basic for numerical evaluation of determinants. (See Section 1.10.) From Rule I1 one deduces that a succession of an even number of interchanges of rows (or of columns) leaves the determinant unchanged, whereas an odd number of interchanges reverses the sign. In each case we end up with apermutation of the rows (or columns) which we term even or odd according to the number of interchanges. From an arbitrary determinant, one obtains others, called minors or the given one, by deleting k rows and k columns. Equations (1.31) and (1.32) indicate how a given determinant can be expanded by minors of the first row. There is a similar expansion by minors of the first column or by minors of any chosen row or column. In the expansion, each element of the row or column is multiplied by its minor (obtained by deleting the row and column of the element) and by f1. The f signs follow a checkerboard pattern, starting with in the top left comer. From three vectors u, v, w in space, one obtains a determinant

+

One has the identities

The vector expressions here are called scalar triple products. The equality D = u . v x w follows from expansion of D by minors of the first row and the formula (1.20) applied to v x w. The other equalities are consequences of Rule I1 for interchanging rows. In (1.34), one can also interchange . and x . For example,

since the right-hand side equals w , u x v. Also, interchanging two vectors in one of the scalar triple products changes the sign:

The number D in (1.34) can be interpreted as plus or minus the volume of a parallelepiped whose edges, properly directed, represent u, v, w as in Fig. 1.12. For where Iwl cos 4 is the altitude h of the parallelepiped (or the negative of h if 4 > n/2), as in Fig. 1.12. Also, lu x vl is the area of the base, so that D is indeed f the volume. One sees that the holds when u, v, w form a positive triple and

+

11

12

Advanced Calculus, Fifth Edition

Figure 1.12

Scalar triple product as volume.

Figure 1.13

Parallelogram formed by u, v .

that the - holds when they form a negative triple. When the vectors are linearly independent, one of these two cases must hold. When they are linearly dependent, the parallelepiped collapses, and D = 0; in the case of linear dependence, either u or v x w is 0, or else the angle 4 is n/2. Thus we have a useful test for linear independence of three vectors, u, v, w in space: They are linearly independent precisely when D # 0. This discussion can be specialized to two dimensions. For two vectors u, v in the xy-plane, one can form

Now u x v = (u,vy

-

uyuX)k= Dk. Thus

D = f l u x v J = &(area of parallelogram),

(1.35)

where the parallelogram has edges u, v as in Fig. 1.13. Again D = 0 precisely when u, v are linearly dependent. We observe that

and hence D is positive or negative according to whether u, v, k form a positive triple. We verify that if 4 is the angle from u to v (measured in the usual counterclockwise sense for angles in the plane), then the triple is positive for 0 < 4 < n and negative forn
I:

(

13

Chapter 1 Vectors and Matrices

We consider a system of three equations in three unknowns:

With this system we associate the determinants

Cramer's Rule asserts that the unique solution of (1.36) is given by

provided that D # 0.

I

1

We can derive the rule by multiplying the first equation of (1.36) by '" a?? a?? (that is, by the minor of a l in D ) , the second equation by minus the minor of a z l ,and the third by the minor of a31.If we then add the equations, we obtain

The coefficient of x is the expansion of D by minors of the first column. The coefficient of y is the expansion of a12 aa32 22

a12

:/j

y:j=o

and similarly the coefficient of ;is 0. The right-hand side is the expansion of D l by minors of the first column. Hence and similarly Dy=D2,

Dz=D3.

Thus each solution x , y , z of (1.36) must satisfy (1.39) and (1.40). If D # 0, these are the same as (1.38); we can verify that, in this case, (1.38) does provide a solution of ( 1.36) (Problem 15). Thus the rule is proved.

14

Advanced Calculus, Fifth Edition If D = 0, then (1.39) and (1.40) show that, if there is a solution, then D l = 0, D2 = 0, 0 3 = 0. We reserve discussion of the general case D = 0 to Section 1.10 and here consider only the homogeneous system

a31x

+ a32y + a3sz = 0.

Here D I = 0, D2 = 0, D3 = 0. Thus if D

# 0, Eqs. (1.41) have the unique solution

called the trivial solution. On the other hand, if D = 0, then Eqs. (1.41) have infinitely many solutions. To show this, we introduce the vectors

Thenxv, +yv2+zv3 h a s c o m p o n e n t s a ~ ~ x + a+al32, ~ ~ y azlx+a22y +a2329 a 3 , x + a 3 ~ y ~332.Hence Eqs. (1.41) are equivalent to the vector equation

+

Now we have assumed that D = 0. It follows that v l , v2, v3 are linearly dependent. For the corresponding determinant, v1 ' v2 x v3 equals D with rows and columns interchanged; by Rule I of Section 1.4, this determinant equals D and thus is 0. Therefore v,, v2, v3 are linearly dependent, and numbers c l , c2,c3 that are not all 0 can be found such that Thus .x = C I I , y = c2t,z = c3t, where t is arbitrary, provides infinitely many solutions of (1.43) and hence of (1.41). The results established here extend to the general case of n equations in n unknowns:

Here

Chapter 1 Vectors and Matrices and D l , . . . , D, are obtained from D by replacing the entries of the first, second, . . . , nth columns of D by k l , . . . , k , as in (1.37).Cramer's Rule holds: If D # 0, then Eqs. (1.44) have the unique solution

If k l = 0 , . . . , k , = 0, then Eqs. (1.44) are homogeneous and always have the trivial solution

If D # 0, this is the only solution; If D = 0, then there are infinitely many solutions of the homogeneous equations. For further discussion of this topic, see Section 1.10.

PROBLEMS 1. Let points P I : ( 1 , 0 , 2 ) ,P.: ( 2 , 1 , 3 ) . P3: ( l , 5 , 4 )be given in space. a ) From a rough graph, verify that the points are vertices of a triangle. b ) Find the lengths of the sides of the triangle. c ) Find the angles of the triangle. d ) Find the area of the triangle. e ) Find the length of the altitude on side Pl P2. f ) Find the midpoint of side PI P2.

g ) Find the point where the medians meet. 2. Let vectors u = i - j 2 k , v = 3i j - k , w = i + 5j

+

a ) Findu . v . C ) Find lul. e ) Find u x (2v 3w). g) Find u x (v x w ) . i) Find comp,v.

+

+

+ 2k be given.

b ) Find u x v. d ) Find ~ ( uv ,) . f ) Findu . v x w andv x u . w . h) Find the direction cosines of u.

3. a ) Show that an equation of the plane through given by

(XI,

y l , z l ) , (x2, y2, z2), (

~ 3 y3. ,

z3) is

Are there any exceptions'? b) Find an equation of the plane through the points of Problem 1 . 4. Let points P I :( 1 , 3, - l ) , P2: ( 2 , 1 , 4 ) , P3: ( 1 , 3 , 7 ) ,P4: ( 5 , 0 , 2 )be given. a ) Show that the points do not lie in a plane and hence form the vertices of a tetrahedron. b ) Find the volume of the tetrahedron.

15

16

Advanced Calculus, Fifth Edition

5. Test for linear independence: a) u = 15i-21k,v=20i-28k b) u=i+j-k,v=2i+j+k,w=7i+5j-k c) u = 2 i + j , v = 2 j + k , w = 2 i + k , p = i + j + k 6. Find parametric equations for the line satisfying the given conditions: a ) passes through ( 2 , 1,O) and ( 3 , 2 , 5 ) . b ) passes through (1, 1 , 2 ) and is parallel to the line x = 2 - 5t, y = 1 2 t , z = 3t. C ) passes through (O,O, 0 ) and is perpendicular to the plane 5x - y z = 2. d ) passes through (1, 2, 2 ) and is perpendicular to the line x = 1 1, y = 2 - t , z = 3 + t and to the line x = 2 + t , y = 5 2 t , z = 7 4t. 7. Find an equation for the plane satisfying the given conditions: a) passes through the z-axis and the point ( I , 2 , 5 ) . b ) passes through (1, 2, 2 ) and is perpendicular to the line x = 2 - r, y = 3 + 5r, z = 2 - 4t. C ) passes through the line x = 1 + r, y = 2 + 3r, z = 1 + 5t and is perpendicular to the plane 2x + y - z = 0. 8. Lct line L I pass through PI and have nonzero vector vl along the line let line L2 pass +' through P2 and have nonzero vcctor vz along the line. Let u = PI P2. Use geometric reasoning for the following: a ) Show that L 1 and L 2 coincide precisely when u x vl = 0 and vl x v2 = 0 b ) Show that L I and L2 are parallel and noncoincident preciscly when u x vl # 0 and V l x v2 = 0. C ) Show that L 1 and L2 intersect at one point precisely when u . vl x v2 = 0 and V l x v2 # 0. d ) Show that Ll and L2 are skewed precisely when u . vl x vz # 0. Show that in this case the shortest distance between two points on L I , L2 is

+

+

+ +

9. Let L be the line through P I :( 1 , 2 , 3 ) and P2: (10. 1 1 , 15).

a ) Find the trisection points of the segment PI P2 b ) Find a point P on L that is not on PI P2 and is two units from P2. 10. Evaluatc the determinants:

11. Dctcrmine whcther the ordcred triple u, v , w is positive or negative:

a) u = i - 2 j - 5 k , v = 2 i + j - k . w = 3 i + 4 j + 2 k b) ~ = 2 i + 3 j + 4 k , v = 4 i + 3 j + 2 k . ~ = i + j + k

+

Chapter 1

Vectors and Matrices

12. Prove the identities:

d) ~ ~ ( V X W ) + V X ( W X U ) + W X ( U X V ) 13. Solve the simultaneous linear equations: a) 3x - 2y = 4, x 2y = 4 b) 5 x - y = 4 , x + 2 y = 3 c) x - y + z = l , x + y - z = 2 , 3 x + y + z = o d) x - y + z = 0 , 3 x - y + 2 z = 0 , 6 x - 4 y + 5 z = O 14. Consider the simultaneous equations

= ~

+

a) Show that if D rically. b) Let D = 0, D l

# 0, Cramer's Rule provides the unique solution. Interpret geomet# 0. Show geometrically why there is no solution.

C) Let D = 0, D l = 0, D2 = 0. Show geometrically various cases that can anisc, some yielding solutions and others yielding no solutions. 15. Show that for the system (1.36) with D # 0, Cramer's Rule (1.38) does provide a solution. [Hint: It suffices to check the first equation of (1.36), since the others are similar. Show that after substitution from (1.38) it can be written a1 1 Dl a 1 2 D 2 a 1 3 D 3= kl D. Show that the left-hand side can be written as

+

+

Now interpret the coefficients of k l ,k?, kg as determinants expanded by minors.] 16. Let D be as in (1.45). a ) Let ai, = a;,, where a,, = 1 for i = j and 6;; = 0 for i # j . Show that D = 1. [This is the rule det I = 1, in the notation of Section 1.9.1 b) Let a,, = 1 if i + j = n + 1 and ai, = 0 otherwise. Show that D = - 1 c) Let ai, = 0 for i > j, so that the array has "upper triangular form." Show that D = a l l a ~ 2 . . . a ,., d ) It can be shown that

where the sum is over all permutations ( j l , . . . , j,,) of (1.2, . . . , n) and F ,,,,,, is 1 for a permutation which is even (obtainable from (1, 2, . . . , n) by an even number of inierchanges of two integers) and is - 1 for an odd permutation (odd number of interchanges). (See Chapter 4 of the book by Perlis listed at the end of the chapter.) Verify this rule for n = 2 and for n = 3.

17

18

Advanced Calculus, Fifth Edition

By a matrix we mean a rectangular array of m rows and n columns:

For this chapter (with a very few exceptions) the objects a l a12, . . . , a,, will be real numbers. In some applications they are complex numbers, and in some they are functions of one or more variables. We call each a;, an entry of the matrix; more specifically, ai, is the ij-entry. We can denote a matrix by a single letter such as A , B, C , X, Y , . . . . If A denotes the matrix (1.48), then we write also, concisely, A = (a;,). Let A be the matrix (aij) of (1.48). We say that A is an m x n matrix. When m = n, we say that A is a square matrix of order n. The following are examples of matrices:

Here A is 2 x 3, and B and C are 2 x 2; B and C are square matrices of order 2. An important square matrix is the identity matrix of order n, denoted by I : 1

for i = j,

0

f o r i # j.

(1.50)

0 0 ...

We call 6;; the Kronecker delta symbol. One sometimes writes I, to indicate the order of I , but normally the context makes this unnecessary. The principal diagonal of a square matrix is formed of the entries a l l , a22, . . . , a,,. For I, the principal diagonal is 1, 1,. . . , 1 (n times). For each m and n we define the m x n zero matrix

One sometimes denotes the matrix by Om, to indicate the size-that is, the number of rows and columns. In general, two matrices A = (ai,) and B = (bij) are said to be equal, A = B, when A and B have the same size and aij = b;; for all i and j . A 1 x n matrix A is formed of one row: A = ( a l l , . . . , a,,). We call such a matrix a row vector. In a general m x n matrix (1.48), each of the successive rows forms a row vector. We often denote a row vector by a boldface symbol: u, v , . . . (or, in handwriting, by an arrow). Thus the matrix A in (1.49) has the row vectors u l = (2, 3 , 5 ) andu2 = (1,2,3).

Chapter 1 Vectors and Matrices Similarly, an m x 1 matrix A is formed of one column:

We call such a matrix a column vector. For typographical reasons we sometimes denote this matrix by col ( a l l ,. . . , a m l )or even by ( a l l ,. . . , a m l ) ,if the context makes clear that a column vector is intended. We also denote column vectors by boldface letters: u, v, . . . . The matrix B in (1.49) has the column vectors vl = co1(1,4) and v2 = col(2, 3). We denote by 0 the row vector or column vector ( 0 , . . . , 0 ) . The context will make clear whether 0 is a row vector or a column vector and the number of entries. The vectors occurring here can be interpreted geometrically as vectors in k-dimensional space, for appropriate k. For example, the row vectors or column vectors with three entries are simply ordered triples of numbers and, as in Section bj ck in 3-dimensional space. This 1.2, they can be represented as vectors ai interpretation is discussed in Section 1.14. Matrices often arise in connection with simultaneous linear equations. Let such a set of equations be given:

+ +

Here we may think of y l , . . . , y, as given numbers and x , , . . . , .u,, as unknown numbers, to be found; however, we may also think of xl, . . . , x,, as variable numbers and y l , . . . , y, as "dependent variables" whose values are determined by the values chosen for the "independent variables" .rl, . . . , x,. Both points of view will be important in this chapter. In either case we call A = ( a i j )the coeficient matrix of the set of equations. The numbers y , , . . . , y, can be considered as the entries in a column vector y = col ( y l . . . . , y,). The numbers x , , . . . , x, can be thought of as the entries in a row vector or column vector x; in this chapter, we usually write X = COI(.YI,. . . , X,).

Let A = ( a i j )and B = ( b i j ) be matrices of the same size, both m x n. Then one defines the sum A B to be the m x n matrix C = ( c i j )such that cij = a,, b,, for all i and j ; that is, one adds two matrices by adding corresponding entries. For example,

+

+

Let c be a number (scalar); let A = ( a i j )be an m x n matrix. Then one defines cA to be the m x n matrix B = ( b i j )such that bij = cai, for all i and j ; that is, cA

19

20

Advanced Calculus, Fifth Edition

is obtained from A by multiplying each entry of A by c. For example,

+

We denote ( - l ) A by - A and B ( - A ) b y B - A. From these definitions we can deduce the following rules governing the two operations: 1. A + B = B + A .

+

+( B + C )= ( A+ B) +C. 4. ( a + b)C = aC + bC. 2. A

+

3. c ( A B ) = c A c B . 5. a ( b C ) = ( a b ) C . 7 . OA = 0 . 9. A + C = B i f a n d o n l y i f C=B-A.

6. I A = A . 8. A + O = A .

(1.54)

Throughout, w e assume that the sizes of the matrices are such that the operations have meaning. T h e proofs are obtained by simply applying the definitions. For example, for Rule 1 w e write

PROBLEMS In these problems the following matrices are given:

1. a) Give the number of rows and columns for each of the matrices A , F, H, L , and P . b) Writing A = (aij),B = (hi;),and so on, give the values of the following entries: a 1 1 , a 2 1 ~ c 2~12~2 ~ d 1 2 ~ fel21 1 ~ ,g 2 3R , Z I , ~ I ~ , ~ B . C) Give the row vectors of C , G, L , and P . d ) Give the column vectors of D, F, L , and N . 2. Evaluate each expression [hat is meaningful: b) C + D. a) A + B . d) L + M. e) N - P .

g) 5 c . j) 2C D

+

-

E.

h) 2 E . k) 3 L - N .

Chapter 1

Vectors and Matrices

3. Solve for X : a) C + X = D ,

4. Solve for X and Y: a) X + Y = N , X - Y = P .

5. Prove each of the following rules of (1.54): a) Rule 2.

b) Rule 3.

c) Rule 4.

e) Rule 6.

f ) Rule 7.

g) Rule 8.

d) Rule 5. h) Rule 9.

1.8 MULTIPLICATION OF MATRICES In order to motivate the definition of the product A B of two matrices A and B, we consider two systems of simultaneous equations:

Such pairs of systems arise in many practical problems. A typical situation is that in which x l , . . . ,x, are known numbers and y , , . . . , y, are sought, all coefficients hi,; and a;,;being known. The second set of equations allows us to compute u I , . . . , u p ; if we substitute the values found in the first set of equations, we can then find y 1, . . . , y, . We carry this out for the general case:

and, in general, for i = I , . . . , m ,

where

,

C..

-a;lbl, -

+ . . . +aipbp, fori

= 1 , . . ..m. j = 1, . . . , n .

(1.57)

Thus from the coefficient matrix A = (ai,)and the coefficient matrix B = (b;,) we obtain the coefficient matrix C = (c,,) by the rule (1.57). We write C = A B and have thereby defined the product of the matrices A and B.

21

22

Advanced Calculus, Fifth Edition

Figure 1.14

Product of two matrices

.

We observe that (aiI , . . . , a;,) is the ith row vector of A and that col (bl, , . . . b,) is the jth column vector of B . Hence to form the product A B = C = (c,,), we obtain each c,, by multiplying corresponding entries of the ith row of A and the jth column of B and adding. The process is suggested in Fig. 1.14. We remark that the product A B is defined only when the number of columns of A equals the number of rows of B ; that is, when A is m x p and B is p x n, A B is defined and is m x n. Also, when A B is defined, B A need not be defined, and even when it is, A B is generally not equal to B A ; that is, there is no cornrnutative law for multiplication. EXAMPLE 1

Herem = 2, p = 2 , n = 3. EXAMPLE 2

Herem = 3 , p = 3 , n = 1. The second example illustrates the important case of the product A v , where A is an m x n matrix and v is an n x 1 column vector. The product A v is again a column vector u, m x 1. In the general product A B = C , as defined above, we note that the jth column vector of C is formed from A and the jth column vector of B , for the jth column vector of C is

Chapter I

Vectors and Matrices

Hence if we denote the successive column vectors of B by u l , . . . , u,, then the column vectors of C = A B are A u l , . . . , Au,. Symbolically,

EXAMPLE 3

To calculate A B , where 3

1

1

2 -1

UI

0

3

2 u2

4 u3

-1 u4

4 us

we calculate

and then note that u4 = Therefore

EXAMPLE 4

U,

and u5 = u?,SO that Au4 = AUI and Aus = Au3.

The simultaneous equations 3x + 2 y $ 5 2 = u , 4x - 5y

-

82 = v ,

7x + 2 y +9z = w

are equivalent to the matrix equation

''

for the product on the left-hand side equals the column vector

[it P i;]

and this equals col ( u , v , w) precisely when the given simultaneous equations hold. In the same way the two sets of simultaneous equations (1.55) and (1.56) can be replaced by the equations Au =y

and

B x = u.

The elimination process at the beginning of this section is equivalent to replacing u by B x in the first equation to obtain y = A(Bx). Our dejinition of the product A B is then such that y = A(Bx) = (A B)x. Therefore for every column vector x,

23

24

Advanced Calculus, Fifth Edition Powers qf a square matrix. If A is a square matrix of order n , then the product A A has meaning and is again n x n ; we write for this product. Similarly, A3 = A 2 A . = A'A, . . . , As+' = A S A , . . . . We also define A0 to be the n x n identity matrix I . Negative powers can also be defined for certain square matrices A ; see Section 1.9. Rules,for multiplication. Multiplication of matrices obeys a set of rules, which we adjoin to those of the preceding section:

20. Ax = Bx

for all x if and only if

A = B.

Here the sizes of the matrices must again be such that the operations are defined. For example, in Rule 13, if A is rn x p, then B and C must be p x n . To prove Rule 10 (associative law), we let C have the column vectors ul , . . . , uk. Then B C has the column vectors B u , , . . . , B u k , and hence A ( B C ) has the column vectors A ( B u I ) ,. . . , A ( B u k ) . But as was remarked above, A ( B x ) = ( AB ) x for every x. Therefore A ( B C ) has the column vectors ( A B ) u l , . . . , ( A B ) u k .But these are the column vectors of ( AB ) C . Hence A ( B C ) = ( AB ) C . For Rule 1 I , A is, say, m x p, and I is p x p. so that

We can also write AI = C = ( c i j ) ,where c,j = ai1S1;

+ . . . + u ; ~ S ~=, a;,,

since S i , = 1 but S;, = 0 for i # j. Rule 12 is proved similarly. For Rule 13 we have A ( B C ) = D , where

+

+

and hence D = A B A C . Rule 14 is proved similarly. Rules 15 and 16 follow from the fact that all entries of 0 are 0; here again the size of 0 must be such that the products have meaning.

Chapter 1 Vectors and Matrices In Rules 17, 18, and 19, A is a square matrix, and k and I are nonnegative integers. Rule 17 is true by definition of A'; and Rules 18 and 19 are true for 1 = 0 and 1 = I by definition. They can be proved for general I by induction (see Problem 4 below). For Rule 20, let A and B be rn x n, and let e l . . . . , e,, be the column vectors of the identity matrix I of order n. Then A 1 is a matrix whose columns are A e , , . . . . Ae,,. and 6 1 is a matrix whose column vectors are B e , , . . . , Be,. If A x = B x for all x , then we have A e , = B e l , . . . , Ae, = Be,.

Therefore A I = B I or, by Rule 1 1, A = B . Conversely. if A = B , then A x = B x for all x , by the definition of equality of matrices. Remark. Because of the associative law, Rule 10, we can generally drop parentheses in multiple products of matrices. For example, we replace [ A ( B C ) ] Dby A B C D . No matter how we group the factors, the same result is obtained. ,

PROBLEMS Let the matrices A , . . . . P be given as at the beginning of the set of problems following Section 1.7. 1. Evaluatc each expression that is meaningful:

a ) AB. d ) C D and DC.

b) CA. e) CE and EC.

g) IL. j) E ( E - 51).

h ) 21

c ) AC. f ) AI. i ) C' - 3 c - l o c o

+ 3GL.

I) MPG. o) OC+ N . r) E ~ .

k ) LNI.

n) KA. 9) E'.

m) HL+ J . p) E?. S) HELP.

2. Calculate R S for each of the following choices of R and S:

3. Consider each of the following pairs of simultaneous equations as cases of (1.55) and (1.56) and express yl, . . ..in terms of . X I , . . . (i) by eliminating u l , . . . and (ii) by multiplying the coefficient matrices: 3ul + 2uz = yl 5x1 - X 2 = U I and 5 ~ 1 6ul = y2 . X I + 2x2 = u2'

{

+

2ui - u2 = ?'I 5 ~ 1 u2 = J'2

+

and

{

X I +2.rz - x s = ul 2x1 3x2 X 3 = ~

+

+

2 '

25

26

Advanced Calculus, Fifth Edition 4. Prove each of the following rules of Section 1.8: a ) Rule 12. b) Rule 14. C ) Rule 18, by induction with respect to I . d ) Rule 19, by induction with respect to 1 and Rule 18. 5. Let A be a square matrix. Prove: a ) A' - I = ( A + I ) ( A - I ) . - I = ( A - I ) ( A+ ~ A + I). b) 2 A 31 = ( A 3I)(A I). C) d ) 6 ~ ~ - A - 21 = ( 2 A + I ) ( 3 A - 2 1 ) . 6. If A and B are n x n matrices, is A' - B' necessarily equal to ( A - B ) ( A + B)? When

+

must this be true? 7. Find nonzero 2 x 2 matrices A a d B such that + B~ = 0 . 8. Prove: If A is a 2 x 2 matnx such that A B = B A for all 2 x 2 matrices B , then A = cI for some scalar r.

Let A be an n x n matrix. If an n x n matrix B exists such that A B = I , then we call B an inverse of A . We shall see below that A can have at most one inverse. Hence if A B = I , we call B the inverse of A and write B = A-'. For a general n x n matrix A we denote by det A the determinant formed from A ; that is,

We stress that det A is a number, whereas A itself is a square array-that is, a matrix. The principal properties of determinants are summarized in Section 1.4. If A and B are n x n matrices, then one has the rule det A det B = det (A B ) .

(1.60)

For a proof, see CLA, Sections 10-13 and 10-14 or page 80 of the book by Perlis listed at the end of the chapter; see also Problem 9 below. From this rule it follows that if A has an inverse, then det A # 0. For A B = I implies det ~ d e Bt = d e t I = 1, so that det A # 0; also det B = det A-' # 0 and, in fact, 1 det B = det A-' = det A '

Chapter 1 Vectors and Matrices Conversely, if det A # 0, then A has an inverse. For if det A # 0, then the simultaneous linear equations

can be solved for xl,

all.rl

+

. ..

+ alnxn = YI

UnlX1

+

...

+

(1.62) UnnXn

= Yn

. . . , xn by Cramer's Rule (Section 1.5). For example,

where D = det A . Upon expanding the first determinant on the right, we obtain an expression of the form

with appropriate constants bll, . . . , bin. In general,

Now our given equations (1.62) are equivalent to the matrix equation Ax = y,

and the solution (1.63) is given by x = By.

The fact that this is a solution is expressed by the relation ABy=y

or

ABy= ly.

This relation holds for all y. Hence by Rule 20 of Section 1.8 we must have A B = I , so that B is an inverse of A . The reasoning just given also provides a constructive way of finding A p l . One simply forms the equations (1.62) and solves for X I ,. . . , x,. The solution can be written as x = B y , where B = A-'. EXAMPLE 1

A =

[::]. The simultaneous equations are 2x1 + 5 ~ 2= y1,

XI

+ 3x2 = Y ? .

We solve by elimination, and find

Therefore A-' =

[ i ] . We check by verifying that A A - '

=I.

27

28

Advanced Calculus. Fifth Edition Nonsinglrlar matrices. A matrix A having an inverse is said to be nonsingular. Hence we have shown that A is nonsingular precisely when det A # 0. A square matrix having no inverse is said to be singrilar. Now let A have an inverse B , so that AB = I . 'Then as remarked above, also det B # 0. so that B also has an inverse B - ' , and BB-' = I . We can now write BA = BAI = B A B B ~ '= B ( A B ) B ~= ' BIB-' = BB--' = I

Therefore, also, BA = I . Furthermore, if AC = I , then

This shows that the inverse,of A is unique. Furthermore, if CA = I , then C=CI=CAB=IB=

B.

These results can be summarized as follows: The inverse B = A-' of A is unique and B satisfies the two equations AB= I

and

BA= I ;

.furthermore, if a n1utri.r B satisfies either one of these two equations, then B must satisjj die other equation. and B = A - ' . The inverse satisfies several additional rules:

Here A and D are assumed to be nonsingular n x n matrices. To prove Rule 2 1, we write

Therefore D - ' A p ' must be the inverse of A D . The proof of Rule 22 is left as an exercise (Problem 5 below). For Rule 23 we reason that A - ' is nonsingular and hence A p l has an inverse. But A - ' A = I , so that A is the inverse of A ' ; that is, A =(Ap1)-'. Rule 2 1 extends to more than two factors, for example, The proof is as above. In this way we see that the product of two or more nonsingular matrices is nonsingular. Negative powers of a square matrix. Let A be nonsingular, so that A-' exists. For each positive integer p we now define A-Qo mean ( A - I ) " . Since A is nonsingular, A'' is also nonsingular; in fact, AQas the inverse ( A A . . . A)-' = A - ~ A - ] .. . A - 1 = ( A - ' ) P .

Therefore

Chapter 1 Vectors and Matrices Rules 18 and 19.

are now satisfied, for A nonsingular, for arbitrary integers p and q, positive, negative, or zero. These rules are proved for p and q nonnegative in Section 1.8. To prove (1.65) for general p and q , we first reason that, since

(1.65) is equivalent to APAqA-P-9 = I. This in turn is equivalent to the statement

(*I

Aq9Ar=I

whenever p

Now for s and t nonnegative we know that A"'

+ q + r = 0. = A"'. This implies

and hence

The last six equations state that (*) holds in each of the following cases: p >_ 0, q 2 0, - 0;p < O , q L O , r > O ; p L O , q i O , r L 0 ; p 2.0, r 5O;p iO,q 5 0 , r > q 5 0, r 5 0; p 5 0, q >_ 0, r 5 0. These are all possible cases. Hence (*) is proved, and (1.65) follows. We note that, by (1.65), APA9 = A9AP. Hence AP and A9 commute (that is, obey the commutative law) under multiplication. The proof of (1.66) is left as an exercise (Problem 6 below). The procedure of Example 1 for finding the inverse of a matrix can be much improved. The calculation of inverses of matrices is very important in numerical work, and various procedures have been devised for this purpose, especially adapted to digital computers (see Section 1.10). Inverses can be used to solve matrix equations-for example, equations of the form A X = B or XA = B , where A and B are known and X is sought; if A is n x n and nonsingular, then we find, respectively, X=A-'B

and

X=BA-'

and verify in each case that the equation is satisfied. As a special case, we solve the equation Ax = y ( A being n x n) to obtain x = A - ' y . If k < n, A is an n x k matrix, and B is a k x n matrix, then A B is an n x n matrix and A B is singular. To see this, we complete A and B to n x n matrices by adding extra columns and rows of zeros. Let A I and B , be the expanded matrices obtained. Then A B = A , B 1 ,since the added zeros contribute nothing to

Remark.

29

30

Advanced Calculus, Fifth Edition the product. But A l and B1 are singular, so that A I B1 = AB is singular, since det A l Bl is 0 by (1.60).

EXAMPLE 2

2

1

5

11

13 35

PROBLEMS 1. Verify that each of the following matrices is nonsingular and find the inverse of each:

2. Let A, . . . , D be as in Problem 1. With the aid of the answers to Problem 1, solve for X or x:

i) xC = (1.2, 3).

j) X D = (2, 1,O).

3. Simplify:

a ) [(AB)-'A-']-'. b) (ABC)-'(C-~ B-IA-I)-I. C) { [ ( A - ] ) * B ] - ~ A - ~ B - ~ ) - ~ . 4. Prove: a ) If A and B are nonsingular and A B = BA, then A-' B-I = B - ' A - I . b) I f A B C = I,then BCA = I andCAB = I. 5. Prove Rule 22. 6. Prove the rule (1.66) for arbitrary integers p, q. [Hint: For s 2 0, t 2 0, show by (1.64) that (AS)-' = (Aps)' = A-St and that (Aps)-' = AS'.]

7. Let A = (a,,) be a nonsingular square matrix of order n. Prove:

<::'].

det A - a 2 , a) Ifn = 2 , t h e n ~ = ' L['22

Chapter 1

Victors and Matrices

b) If n = 3, then A - - is the matrix

1 det A

-

C) Let Aij denote the minor determinant of A obtained by deleting the ith row and jth column from A. Let hi, = (-]);+'A ji. The matrix B = (bij) is called the adjoint of A and is denoted by adj A. Show that 1 A-I adj A. det A 8. Let all matrices occurring in the following equations be square of order n. Solve for X and Y , stating which matrices are assumed to be nonsingular:

9. a ) Prove the rule (1.60) for 2 x 2 matrices; that is, prove:

[Hint: Use Rule V of Section 1.4 for adding determinants to write the right-hand side as a sum of four determinants. Then factor bl 1. . . . from these and show that two terms are 0 and the other two have sum

b) Prove the rule (1.60) for 3 x 3 matrices. C) Prove the rule ( 1.60) for n x n matrices. 10. Let A = col (U1 , . . . . u,,) (column vector) and B = (vl , . . . , v,) (row vector); let n 2 2. a ) Show directly that thc n x n matrix AB is singular (cf. the Remark at the end of Section 1.9). b) Show that BA is 1 x I- that is, a scalar. 11. (Permutation matrices) Let Pij be the n x n matrix obtained from I by interchanging the ith and jth rows (i < j). Thus for n = 3,

a ) Show that P; = I. b) Show that Pij can be obtained from I by interchanging the ith and jth columns of I. c) Show that, if A is an n x n matrix, then PjjA is obtained from A by interchanging the ith and jth rows of A, and A Pij is obtained from A by interchanging the ith and j th columns of A. d ) Show that Pij is nonsingular.

31

32

Advanced Calculus, Fifth Edition

For solving simultaneous linear equations, finding inverses of matrices, and evaluating determinants, a number of practical techniques have been developed and implemented in computer software. We here summarize the standard Gauss and Gauss-Jordan procedures and refer for details to the book of Gerald and Wheatley and that of Perlis, listed at the end of the chapter. Gauss procedure. The system (1.44) of Section 1.5 is replaced by the matrix equation Ax = k,

(1.67)

where A = (ai,) and k = col (k, , . . . , k,). One forms the "augmented matrix" formed from A by adjoining k as (n

+ 1)-st column. One operates on the matrix B by

I. adding multiples of one row to another row, 11. interchanging two rows. When A is nonsingular, suitable applications of these steps replace B by a new matrix C = [D

h],

(1.69)

obtained by augmntation from an upper triangular matrix D-that is, a matrix with 0's below the principal diagonal. The steps described correspond to replacement of the given simultameous equations (1.44) by new ones with the same solutions. Since D is upper triangular, the new equations have the form

.

and are easily solved for x,,, x , - ~ . . . . x, in turn. Thus we have found the unique solution. We observe that by Rules I1 and VI of Section 1.4, det D = ( - l ) e det A ,

(1.71)

where C is the number of row interchanges. Thus, for nonsingular A , det A and det D . . . dnn # are both nonzero. By Problem 16 following Section 1.5, det D = d l 0, . . . , as is needed to solve (1.70), and by (1.71) one can at once obtain det A. Gauss-Jordan procedure. In addition to steps I and 11, we use a step:

111. multiplying a row by a nonzero scalar. By applications of these three steps we can obtain from the matrix B a new matrix E=[l

PI,

and the new equations (still having the same solutions) have the form

we have again obtained the unique solution.

(1.72)

Chapter 1 Vectors and Matrices The inverse of A can be obtained by the same procedure, applied to the matrix F = [A

I].

(1.74)

obtained from A by adjoining to it the successive columns of the identity matrix of order n. The steps described can be applied to F to reduce it to the form

and one verifies that H = A p ' (Problem 9 below). Singular case. When A is singular, one cannot proceed from (1.68) to (1.72). However, one can always use steps I and I1 to reduce B to the form

where K is an n x n matrix in "row echelon form." By this we mean that in each row the first nonzero entry (the "leading entry") has column index which increases as the row index increases. An example is

In the singular case. the first r rows (0 5 r 5 n - 1 ) contain nonzero entries, whereas the last n - r rows are all 0; here r is the rank of the matrix. In the nonsingular case r = n. By steps I, 11, and 111. one can in fact achieve a form (1.76) in which K has row echelon form and each leading entry is 1 and is the only nonzero entry in its column. The first r of the corresponding linear equations can now be solved for r of the unknowns in terms of thc rcrnaining n - r unknowns, which can be chosen arbitrarily. These yield infinitely many solutions, provided that the remaining n - r equations contain no contradictions. Since the last n - r rows of K are zero, one must have c,.,, = cr+?= . . . = c,, = 0. Otherwise there is no solution. If the given simultaneous equations are homogeneous, then the steps described preserve that property, so that c must be 0 and no contradiction can arise. Thus if det A = 0, so that A is singular; the simultaneorrs equations A x = 0 have infinitely many solutions. In the extreme case r = 0, A must be the zero matrix and there is no solution if k # 0, all x are solutions if k = 0.

PROBLEMS 1. Solve by Gausslan elimination:

a) x + v - ; = I. 2 r + y + z = 5 , 2x+y=O b) x + 2 y + z = 4 . 2.1 - ~ + z = 4 . x - y + 2 z = O c) y + 2 z = 3 . -I + y - ; = 3 , 2x+z=4 d) 2.1 + y + 3 : = 6 , 2x+y+Sz=8, -2x+3v-4:=-3

33

34

Advanced Calculus, Fifth Edition

2. Evaluate the determinant as in Gaussian elimination:

3. Solve by Gauss-Jordan elimination:

4. Find the inverse matrix:

5. In the following systems the elimination process shows that the determinant of coefficients is 0. Carry out the process and find all solutions:

a) 2 x - y + z = 3 , ~ + 2 y - z = 1 , 5 ~ - 5 y + 4 ~ = 8

6. Consider two equations in three unknowns:

a) Show that if Gaussian elimination can be carried out to solve for x and y, then by writing z = t the solutions become parametric equations for a line in space (Section 1.3). b) Assume that the two equations represent two planes in space. Interpret geometrically the case in which the equations have no solution and the case in which elimination leads to a second equation 0 = 0. 7. Consider four equations in three unknowns:

Assume that the pair of equations for i = 1 , 2 and the pair for i = 3 , 4 each represents a line in space as in Problem 6(a). Thus the solutions of the four equations represent the points common to two lines in space. Discuss the geometrical alternatives that can occur and relate them to the set of solutions of the given equations. 8. Let A be an m x n matrix. Let C = (cij) be the m x m matrix which differs from the identity only in that c h h = A # 0. Let B be the m x m matrix which differs from the identity only in the kth column, in which bkk = 1. a) Show that CA is obtained from A by multiplying the hth row by A. b) Show that BA is obtained from A by adding bjk times the kth row to the jth row for all j except k. C)

Show that B and C are nonsingular.

Chapter 1 Vectors and Matrices Remark These results show that steps 1 and 111 of Section 1.10 are obtained by multiplying A on the left by a nonsingular matrix. The same assertion applies to step I1 (Problem 11

following Section 1.9). 9. Let AH = I and let H have column vectors k l . . . . . k,, so that Ak, equals q,, the jth column of I. Show that the Gauss-Jordan process of going from (1.74) to (1.75) is

equivalent to solving the n systems Ak, = q, to obtain H.

Let A be an n x n matrix. For some nonzero column vector v = col ( v l , . . . . v,) it may happen that, for some scalar A , Av = hv.

(1.78)

If this occurs, we say that h is an eigenvalue of A and that v is an eigenvector of A, associated with the eigenvalue A . The concept of eigenvalue has important applications in many branches of physics. An important example is the spectrum-f light, of an atom, of a nucleus. The frequencies occurring in the spectrum correspond to the eigenvalues of a matrix (or of a suitable generalization of a matrix). Eigenvalues are also of importance in the solution of linear differential equations (see Chapters 9 and 10). We can write Eq. ( 1.78) in the form Av = h l v or in the form

Thus v = col (v,, . . . , v,) is an eigenvector of A associated with the eigenvalue h precisely when v l , . . . , v, form a nontrivial solution of the set of homogeneous linear equations

Now we know (Section 1.10) that Eqs. ( 1.79) have a nontrivial solution precisely when the determinant of the coefficients is 0, that is, when det (A - h l ) = 0 or

When expanded, ( 1 2 0 ) becomes an algebraic equation of degree n for A , called the characteristic equation of the matrix A. The eigenvalues of A are simply the real roots of the characteristic equation. (The complex roots of the characteristic equation can also be interpreted as eigenvalues of A; see the end of this section).

35

36

Advanced Calculus. Fifth Edition EXAMPLE 1

Let A =

[:i].

Then the characteristic equation is

The roots are h l = 4 and hz = - 1. To find v for h = A I = 4, we form the equations

and find that the solutions are all vectors k(2, 3). Hence the eigenvectors associated with the eigenvalue h l = 4 are all vectors k(2.3) for nonzero k. In the same way we find that all eigenvectors associated with the eigenvalue A? = - I are all vectors k( I . - 1 ) for nonzero k. EXAMPLE 2

Let B = (h,6,,) = 0

0

...

A,,

We call B a diagonal matrix and write B = diag(hl. . . . , A,]). Then B has the characteristic equation

Hence B has the eigenvalues h l , . . . , A,,. We leave to Problem 5 below the discussion of the eigenvectors of B. We remark here that when h , . . . . h,, are n different numbers, the eigenvectors associated with the eigenvalue hk are the vectors ((0, . . . . 0 , 1.0. . . . , O), with 1 as kth entry and c nonzero.

.

Sirrlilur matrices.

Let A and B be n x n matrices. We say that B is similur to A if

for some nonsingular n x n matrix C . If this holds, then

so that A is also similar to B. Hence we speak of similar matrices A. B. IfA and B are similar, then A and B have the same characteristic equation. For det(B - h l ) = d e t ( C - ' ~ C- h l ) = d e t ( ~ - ' ~~ c = det C - I

c-~Ic)

det ( A - Al)det C = det (A - h l ) ,

since det C-I det C = 1. It follows also that A and B have the same eigenvalues Matrices with n distinct real eigenvalues. Let the tz x n matrix A have n distinct (real) eigenvalues h l , . . . , A, and let v l , . . . v,, be eigenvectors associated with hi, . . . . A,, respectively: Av, = h;v,, i = 1 . . . . , n . Now let C be the matrix whose

.

Chapter 1 Vectors and Matrices column vectors are vl, . . . , v,, respectively. Write v, = col ( v l j , . . . , vnj) for j = I , . . . , n.Then

... -

= CB, U,I

" '

...

An

where B is a diagonal matrix, B = diag ( A l , . . . , A,). It can be shown that C must be nonsingular (see Problem 12 following Section 1.16). Hence B = C-'AC, and A is similar to the diagonal matrix diag ( A , , . . . , A,).

EXAMPLE 3

Here A has the characteristic equation

For A = 3 the eigenvector v = (vl, v2, u3) must satisfy the equations

Hence vl = ( 1 , 0 , 1) is an associated eigenvector. Similarly, for ,I2 = 4 an associated eigenvector is (2,2, l), and for A3 = 5 an associated eigenvector is (0, 1, - 1).Hence C-'AC = B, with

We check by verifying that AC = CB. Thus A is similar to the diagonal matrix B. In general, some of the eigenvalues may coincide (multiple roots of the characteristic equation), and there may be less than n real roots of the characteristic equation (1.80). To handle the general case, it is best to extend the theory of matrices to the case of complex matrices, whose elements are complex numbers. All concepts generalize easily to this case, and, in particular, one can define complex eigenvalues and complex eigenvectors as above. For example, the matrix

37

Advanced Calculus. Fifth Edition

+

has the characteristic equation h2 1 = 0. T h e eigenvalues are 33,and associated eigenvectors are (2, 1 7 i). O n e proves, exactly as above, that when the matrix A has n distinct complex eigenvalues, A is similar 10 a diagonal matrix.

PROBLEMS 1. Find all eigenvalues and associated eigcnvectors:

b ) c ( d For each of the matrices of Problem I , call the matrix A and find a nonsingular matrix C and a diagonal matrix B such that B = C - ' A C .

2. a

3. (Comp1e.r cctse) For each of the tollowlng choices of rnatr~xA, find all eigcnvalues and associated eigenvectors, and find a diagonal matrix B such that B = C-' AC:

4. (Repp~~tpd rools) Find all cigenvalues and associated eigenvectors:

5. (Diagonal matrices)

a) Let B = diag(A. p ) . Show that, if A # g ,thcn the eigenvcctors associated with A. are all nonzero vectors c( 1, O j and those associated with p are all nonzero vectors c(0, 1); show that, if h = p, then the eigenvectors associated with h are all nonzero vectors (111, UZ). b) Let B = diag ( A [ , A2, A 3 ) and let el = ( 1 , 0 , 0 ) .e2 = (0, 1. O), e3 = (0.0. 1 ) (column vectors). Show that if A l . A.2, A3 aredistinct, then for each Ak the associated eigenvectors are the vectors cek for c # 0: show that if hi = A? # A3, then thc cigenvectors associatcd with Al are all nonzero vectors r i e l c ~ e 2and thosc associated with A 3 are all nonzero vectors ce3: show that if A1 = A] = A3, then the cigcnvectors associated with A l are all nonzero vectors v = ( v l v l , v3). c ) Let B = diag ( A l , . . . , A,). Show that the eigenvectors associatcd with the eigenvalue hi are all nonzero vectors v = ( u l , . . . . v,,) such that v, = 0 for all i such that A, # h k .

+

6. Let A and B besimilarmatrices. Let A I , . . . , A,, be the(not necessarily distinct)eigenvalues of A. a) Prove that h 1 A.? . . . A,, = det A b) Concludc from the result of (a) that det A = det B C) Provethati! + . . . + A , = a l l + . . . + a , , , , . T h e n u n l b e r a l ~ + . . . +a,,, iscalledthe trace of A. d) Prove from the rcsult of (c) that A and B havc equal traces. [ I ] is not similar to a diagonal matrix. [Hint: Show by the 0 .I results of Prohlcm h that the d~agonalmatrix would have to be I.]

7. Prove that the matrix A =

Chapter 1 Vectors and Matrices 8. Prove the following: a) Every square matrix is similar to itself. b) If A is similar to B and B is similar to C, then A is similar to C.

*1.12 THETRANSPOSE Let A = (a;,) be an m x n matrix. We denote by A' the n x m matrix B = ( b i j )such that hi, = a,; for i = 1 , . . . , n , j = 1, . . . , m . Thus B = A' is obtained from A by interchanging rows and columns. The following pair is an illustration:

The first row of A becomes the first column of A'; the second row of A becomes the second column of A'. In general, we call A' the transpose of A. We observe that I' = I . The transpose of a matrix obeys several rules, which we adjoin to our list: 24. ( A

+ B)' = A' + B ' .

25. (cA)' = c A ' . 26. (A')' = A. 27. ( AB)' = B'A'. 28. If A is nonsingular, then ( A p ')' = (A1)-' To prove Rule 24, we write D = ( A + B ) = ( d i j ) ,so that

for all i and j. Then D' = E = (e,,). where eIj = d j ; for all i and j, or pi.;

+

= aJi

+ b,,.

+

Thus E = A' B' or D' = A' B ' . The proofs of Rules 25 and 26 are left as exercises (Problem 4 following Section 1.13). To prove Rule 27. we let A be m x p, B be p x n. We then write C = A B = (c,,), D = A' = (d;,), E = B' = (e;,). Then B'A' = ED =

F = (f,,),

where

,, - e,ldl, + . . . +ejpd,;

f . -

,

= bliajl

=~,lb[,+...+a~,b,,~ =c,;

Hence F = C' or B'A' = ( A B ) ' .

+ . . . +b

P Ia JP

( i = 1, . . . , n % j = 1 , . . . , m ) .

39

40

Advanced Calculus, Fifth Edition To prove Rule 28, we write AA-' = I. Then by Rule 27,

From this equation it follows, as in Section 1.9, that (A')-' = ( A p ' ) ' . A matrix A such that A = A' is called a symmetric matrix. Here A must be a square matrix. The matrix I is symmetric, as are the following matrices:

L

Also, every diagonal matrix is symmetric. Symmetric matrices are useful in discussing quadratic forms, that is. algebraic expressions of the form

For n = 2 the expression is

Here ~ 1 x is2 the same as ~ 2 . ~ SO 1 , we could combine the second and third terms. However, it is preferable to split the combined term into two equal terms, each having as coefficient the average of a12 and a2l. For example, 3x: 5 x 1 ~7x2.xI ~ 4.xf is replaced by

+

+

+

By proceeding similarly for the general quadratic form (1.82)we can always assume that the coeficient matrix ( a I j )is symmetric, and it is standard practice to write quadratic forms in this way. For the example just considered, this matrix is Now let the n x n symmetric matrix A = ( a i j )be given, and consider the quadratic form ( 1 3 2 ) . Here we can consider x l , . . . , xn as variables. For each assignment of numerical values to X I , . . . ,xn the form (1.82)has a numerical value Q. Hence Q is a function of the n variables x l , . . . , x,. However, it is simpler to think of Q as a function of the vector x = col ( X I , . . . , x,):

[i:I.

Furthermore, for each x we can compute the number Q(x) by matrix multiplications:

Here x' is the transpose of the column vector x and is therefore the row vector . . . , x , ) . Accordingly, (1.85)is the same as

(XI,

Chapter 1 Vectors and Matrices The product of the last two factors is an n x 1 column vector whose ith entry is ailxl . . . a,,x,. The product of the 1 x n row vector (XI,. . . , x,) and this n x 1 column vector is a 1 x I matrix-that is, a number; in fact, it is precisely the number on the right-hand side of (1.84). Therefore (1.85) is indeed another way of writing (134). As an illustration, we write the quadratic form Q(x) of (1.83) as follows:

+ +

When expanded, this becomes

as expected.

"1.13 ORTHOGONAL MATRICES Let A be a real n x n matrix. Then A is said to be orthogonal if AA' = I.

(1.86)

Hence, A is orthogonal if and only if A - ' = A', that is, if and only if the inverse of A equals the transpose of A . Thus every orthogonal matrix is nonsingular. The following are examples of orthogonal matrices:

Let us consider the row vectors u l , u2 of A as vectors in the xy-plane:

Then we observe that u l and u2 are both unit vectors and that u l . u2 = 0, so that ul, u2 are perpendicular. A similar statement applies to the column vectors of A:

We can proceed similarly with the row vectors or column vectors of B , regarding them as vectors in space: u l = f i $j ik,. . . . Again we verify that the row vectors, or the column vectors, are mutually perpendicular unit vectors.

+ +

41

42

Advanced Calculus, Fifth Edition The geometrical concepts used here can be generalized to n dimensions (Section 1.14). Here we phrase them algebraically. For an n x n matrix A = (ai,) the crucial conditions are as follows: a"...+aa' I I ailajr

Ill

= 1,

+ . . . +ai,~,;,= 0 , a f , + . . . + a ".I' = 1.

i = I . . . . . n.

i

# j. i

= I,...,

(1.87)

.

j = I , . . n,

j = 1, . . . . n ,

a ~ j a l k $ . . . $ a ~ ~ ~ l , ~ ~ j=#Ok .,

j = l ,..., n, k = I

(1.88) (1.89)

. . . . .n. (1.90)

(Here (1.87) states that the row vectors are unit vectors, (1.88) states that different row vectors are orthogonal, and (12 9 ) and ( 1.90) express the analogous conditions on the column vectors.) THEOREM

a) Let A be an n x n orthogonal matrix. Then conditions ( I 2 7 ) through (1.90) all hold. b) If A is an n x n matrix such that (1.87) and (1.88) hold or such that (1.89) and ( 1.90) hold, then A is orthogonal. Proof:

a) Let A be orthogonal, so that (1.86) holds. By the definition of matrix multiplication, (1.86) asserts that

Hence (1.87) and ( 1 3 8 ) follow. From (1.86) we have also, by the properties of inverses (Section 1.9), A ' A = 1.

(1.92)

and hence

and this implies (1.89) and (1.90). b) If (1 3 7 ) and (1.88) hold, then (1.91) holds, so that A A ' = I; if (1 39) and (1.90) hold, then (1.93) holds, so that A'A = I. In either case, A is orthogonal. Orthogonal matrices are important in studying changes of coordinates (see Problem 13 following Section' 1.15). If two n x n matrices B , C are such that

for some orthogonal matrix A , then B is said to be orthogonally congruent to C . It then follows that C is also orthogonally congruent to B, so that we refer to B, C

Chapter 1 Vectors and Matrices as orthogonally congruent matrices (Problem 9 following this section). We observe that orthogonally congruent matrices are similar. IfC is symmetric, then C is orthogonally congruent to a diagonal matrix B, B = diag ( A I , h2, . . . A,), where h l , . . . , h, are the (necessarily real) eigenvalues of C (with multiple eigenvalues repeated in accordance with their multiplicities). For a proof of this theorem, see Section 5-3 of the book by Perlis listed at the end of the chapter (see also Problems 12 and 16 following this section). This result is very important for the study of quadratic forms. Let C be a symmetric matrix and let A be an orthogonal matrix such that A-'CA = B = diag ( A l , . . . , A,). Then let us consider the quadratic form

.

We express x l , . . . , x, in terms of new variables y l , . . . , y, by writing y = c o l ( y l , . . . , y,,).

x=Ay.

Then x' = y'A' = Y'A-', and hence Q(x) becomes a quadratic form Ql(y):

Hence Q(x) can be written in terms of y l , . . . , y, a s a quadratic form containing only the squares of the unknowns. The coefficients h I , . . . , A , are the eigenvalues of B. But B and C have the same eigenvalues. Hence once we know the eigenvalues of C (with their multiplicities), we can write Q in terms of y l , . . . , y,. It is not necessary to find the matrix A. We can interpret x l , . . . , x, as Cartesian coordinates in n-dimensional space (Section 1.14). Then yl , . . . , y, are simply new Cartesian coordinates in n-dimensional space with the same origin (see Problem 14 following Section 1.15). For n = 2 the equation Q(x) = I-that is, the equation

-represents a conic section in the xlx2-plane.In terms of the new coordinates yl , yz this equation becomes Q (y) = 1, or

and hence represents an ellipse or hyperbola. In fact, in the language of plane analytic geometry the theorem on symmetric matrices for the case n = 2 is equivalent to the familiar statement that every second-degree equation AX^ Bxy Cy2 = 1 can be reduced to the standard form Ax2 B~~ = 1 by an appropriate rotation of axes in the xy-plane. (If h l h 2 = 0, the curve is degenerate.)

+

+

+

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Advanced Calculus, Fifth Edition

PROBLEMS 1. Find the transpose of each of the matrices:

2. Choose a and b so that each of the following matrices becomes symmetric:

3. For each of the following quadratic forms, obtain the coefficient matrix when the form is written in such a way that the coefficient matrix is symmetric: a ) 5x: 4 ~ 1 x 2 3x22.

+ + b) 7x: + 2 ~ ~ x x2 i . C) x t + 3x22 x-: + 4 ~ 1 x 2+ 6 x 1 + ~ 2~~ 2 x 3 , d) 2 ~ + : xi+ + 2 x 1 ~4 ~ 2 x 3 . -

-

X:

-

4. Prove each of the following parts of (131). a ) Rule 25,

b) Rule 26.

5. Show that each of the following matrices is orthogonal: cos w b, [ s i n w

sinw cosw

I

6. (a), (b), (c). Represent the row vectors of each of the matrices in Problem 5 (a), (b), (c) as vectors in the plane or in space, graph them, and verify that they are mutually perpendicular unit vectors. 7. (a), (b), (c). Proceed as in Problem 6 with column vectors instead of row vectors. 8. Let A and B be n x n orthogonal matrices. Prove: a ) d e t A = f1. b) AB is also orthogonal. C) A' and A-' are orthogonal. 9. Prove: a ) Every square matrix is orthogonally congruent to itself. b) If B is orthogonally congruent to C, then C is orthogonally congruent to B. C) If B is orthogonally congruent to C and C is orthogonally congruent to D, then B is orthogonally congruent to D.

Chapter 1

Vectors and Matrices

10. We consider complex matrices. that is, matrices with compIex entries. If A = ( a j j )is such a matrix, we denote by A the conjugate of A , that is, the matrix ( ~ 5 ; where ~ ) . ? is the conjugate of the complex number z (for example, if z = 3 + 5i, then 2 = 3 - 5 i ) . Prove the following:

+

+

a) If z l and 12 are complex numbers, then ( z I z 2 ) = ? I 2 2 and zlzz = 7 172. -b) -If A and B are matrices and c is a complex scalar, then ( A B ) = A B , AB = A B , ( c A ) = FA. C) Every complex matrix A can be written uniquely as A 1 i A 2 ,where A l and A? arc real matrices, and A I = + ( A+A),A? = ( 2 i ) - ' ( ~ A ) . We call A l the real part of A. A2 the imaginary pariof A.

+

+

+

-

d ) If A is a square matrix, then (A)'= (A') and, if A is nonsingular, then ( A p ' )= ( A ) - ' .

e) A is a real matrix if and only if A = A . 11. Let A be a real square matrix, let h be real, and let Av = hv for a nonzero complexcolumn vector v. Show that Au = hu for a real nonzero vector u , so that h is an eigenvalue of A considered as a real matrix. [Hint: Let v = p i q , where p and q are real and not both zero. Show, with the aid of the results of Problem 10, that A p = hp and Aq = hq and hence that u can be chosen as one of p, q . ]

+

12. Let A be a real symmetric matrix. Show, with the aid of the results of Problem 10, that all eigenvalues of A are real. [Hint: Let Av = hv for some complex h and some nonzero complex vector v = col ( t i l . . . . , v,). Consider the product Q = vlAC. Show that Q is real and that

Conclude that h is real.]

Remarks By Problem I 1. h is actually an eigenvalue of A as a real matrix. The expression v'AC is a special case of a Hermitian quadratic form Q ( v ) .See Section 5-7 of the book by Perlis listed at the end of the chapter. 13. Let A be as in Problem 12 and let Q ( x ) = x'Ax be the corresponding quadratic form. We call this form positive definite if Q ( x ) > 0 for x # 0. Show that, if Q ( x ) is positive definite, then det A > 0. [Hint: Use Problem 6 following Section 1.1 I .]

Remark It can be shown that Q is positive definite if and only if det Ak > 0 for k = 1 , . . . , n , where Ak = ( a i j ) ,i = 1 , . . . k , j = 1, . . . , k. See Chapter 6 of G. Strang. Linear Algebra and Its Applications, 3rd ed., Int'l Thomson Publishing, 1988.

.

14. Let A and Q be as in Probl,em 13. Show that Q ( x ) can be written as u: + . . . + u i in terms of appropriate coordinates ( u 1 , . . . , u , ). where x = Du and D is nonsinguIar. 15. The inner product of u = c o l ( u ~ ,. .. , u,) and v = c o l ( v l , . . . , v,) is defined in Section 1.14 as u 1 vl + . . . + u,v, and hence equals u'v = v'u. Also lul is defined as (u'u)'"; u and v are termed orthogonal if u'v = 0. Now let A be a real symmetric n x n matrix, so that A = A'. Let h I , h z be eigenvalues of A which are unequal and let v l , V? be corresponding eigenvectors. Show that vl and v2 are orthogonal. [Hint: First show that x'Ay = ylAx for all x, y. Then from A v l = h l v l ,Av2 = h 2 v z deduce that h l v ; v l = h2v',v2 and hence that v',v2 = 0.1 16. For A as in Problem 15, let A have n distinct eigenvalues i l . . . . , A,. Show that A is orthogonally congruent to B = diag ( h l . . . . , A,). [Hint: As in Section 1.1 1, B = C-' AC, where the columns of C are eigenvectors v , . . . . v , of A. From the results of

.

45

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Advanced Calculus, Fifth Edition Problem 15 show that the v, are pairwise orthogonal and can be chosen so that Iv, I = 1 for ,j = 1. . . . , n . ]

1.14 ANALYTIC GEOMETRY AND VECTORS IN n-DIMENSIONAL SPACE The formal operations on vectors and coordinates in Section 1.2 suggest that the restriction to 3-dimensional space, and hence to three coordinates or components, is not necessary. We define Euclidecln n-dimensional space E n to be a space having n coordinates xl, . . . ,x,. Throughout this section, n will be a fixed positive integer. Apoint P of the space is by definition an ordered n-tuple ( . T I , . . . , x,); all points of the space are obtained by allowing x l , . . . , x, to take on all real values. The point (0, . . . , 0 ) is the origin, 0 . The distance between two points A: ( a I .. . . , a,) and B: ( b l ,. . . b,) is defined to be the number

.

A vector v in n-dimensional space is defined to be an ordered n-tuple ( v l , . . . , v,) of real numbers; v , , . . . , v, are the components of v (with respect to the given coordinates). In particular, we define a zero vector

To the pair of points A , B in that order corresponds the vector

2 = ( b )- 0 1 , .

. . . bl, - a,,).

(1.97)

Remarks. Both points and vectors are represented by n-tuples, but this will be seen to cause no confusion. In fact, there are advantages in being able to go back and forth freely between the two points of view, that of p o i n t s ~ n dthat of vectors. With each point P: (x,, . . . , x,) we can associate the vector OP = ( x l , .. . . .u,) = x. Conversely, to each vector x = (xl , . . . , x , , w e can assign the point P whose coordinates are (x,, . . . , x,), and then x = OP. Later we shall also interpret our vectors as matrices, either as row vectors or as column vectors. The sum of two vectors, multiplication of a vector by a scalar, and the scalar product (or innerproduct or dotproduct) of two vectors are defined by the equations:

The length or norm of v = ( v l , . . . , v,) is defined as the scalar

If Ivl = I , v is called a unit vector.

Chapter 1 Vectors and Matrices The vector product of two vectors can be generalized to n dimensions .only with the aid of tensors (see, however, Section 11-9 of CLA). The set of all vectors (vl, . . . , v,) with the operations (1.98) through (1.101) will be denoted by V n . In V n the following properties can then be verified (Problem 6 following Section 1.15): I. u + v = v + u . 11. ( u v ) w = u ( v w). 111. h ( u V ) = h u hv. IV. (a b ) u = a u bu. V. (ab)u = a(bu). VI. lu = u . VIII. u . v = v . u . VII. ou = 0. IX. ( u + v ) . w = u . w + v . w . X. ( a u ) . v = a ( u . v ) . X I . u . u 2 0. XII. u . u = 0 if and only if u = 0. (1.102) A set of k vectors v l ,. . . , vk of V n is said to be linearly independent if an equation

+

+

+ + +

+

+ +

can hold only if c l = . . . = ck = 0. If a relation (1.103) does hold with not all c's equal to 0, then the vectors are said to be linearly dependent.

Remarks. We observe that for every n x k matrix A , Acol(ci, . . . , c ~ ) = c I v I + . . . + c ~ ~ ~ ,

(1.104)

where v i , . . . , vk are the column vectors of A . For

Thus the linear combinations of v l , . . . , vk can be expressed as A c . Accordingly, to state that v l , . . . , vk are linearly independent is the same as to state that A c = 0 is satisfied only for c = 0. In particular, for k = n , A is a square matrix, and A c = 0 is equivalent to n homogeneous linear equations in n unknowns; thus the column vectors v l , . . . , v, of A are linearly independent precisely when these equations have only the trivial solution c l = 0, . . . , c, = 0-that is, when det A # 0, or A is nonsingular. Accordingly, n vectors v i , . . . , v, of V nare linearly independent ifand only $A is nonsingular, where A is the matrix whose column vectors are vl , . . . , v,. A set of k vectors v l , . . . , vk of V n is said to be a basis for V n if every vector v of V n can be expressed in unique fashion as a linear combination of v l , . . . , vk, that is, if for unique choices of the scalars cl , . . . , ck. We call these scalars the components of v with respect to the basis v , , . . . , vk We can now state a number of rules concerning linear independence and basis:

a) The vectors v l , . . . , vk are linearly dependent if and only if one of these vectors is expressible as a linear combination of the others.

47

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Advanced Calculus, Fifth Edition b) If one of the vectors vl , . . . , vk is 0, then v l , . . . , vk are linearly dependent. c) If v l , . . . , vk are linearly independent, but v l , . . . , vk, v k + are ~ linearly dependent, then v k + is ~ expressible as a linear combination of v l , . . . , vk. d) If vl , . . . , vk are linearly independent and h < k, then vl , . . . , vh are linearly independent. e) (Rule for comparing coefficients). If vl , . . . , vk are linearly independent and

then a1 = bl, a;, = b2,. . . ,ak = bk. f ) There exist n linearly independent vectors in Vn: for example, the vectors

(4

g) There do not exist n + 1 linearly independent vectors in Vn. h) If v l , . . . , vn are linearly independent vectors in Vn, then v l , . . . ,v, form a basis for Vn; in particular, e l , . . . , e,, form a basis for Vn. i) Every basis for Vn consists of n linearly independent vectors. j) If k < n and vl , . . . , vk are linearly independent, then there exist vk+], . . . , v, such that v , , . . . , v, form a basis for Vn. k) If v l , . . . , vk are linearly independent vectors in Vn, and u l , . . . , uk+l are all expressible as linear combinations of v l , . . . , vk, then u l , . . . , uk+l are linearly dependent.

We prove Rules (g) and (i) here and leave the remaining proofs to Problem 10 following Section 1.15.

~ linearly independent vectors in Vn.Then by Rule Proof of (g). Let v l , . . . , v n + be (d), vl , . . . , vn are also linearly independent. Hence the matrix A, whose columns are vl ,. . . . , v,, is nonsingular. Therefore the equation Ac = v n + has ~ a unique solution for c; that is, By Rule (a), v l , . . . , vn+]would then be linearly dependent, contrary to assumption. Therefore there cannot be n 1 linearly independent vectors in Vn.

+

Proof of (i). Let vl , . . . , vk be a basis for V". Then vl , . . . , vk are linearly independent. For if c l v l . . . ckvk = 0, then

+ +

since, by definition of a basis, 0 can be represented in only one way as a linear combination of v l , . . . , vk.Hence by Rules (d) and (g) we must have k _( n. Let us suppose k < n. Then we can express e l , . . . , en as linear combinations of vl, . . . , vk-say, Abl = e l , . . . , Ab,, = en, where A is the n x k matrix whose column vectors are vl, . . . , vk.Accordingly, A B = I, where B is the k x n matrix whose column vectors are b,, . . . , b,. But fork < n a product of an n x k matrix and a k x n matrix

Chapter 1

Vectors and Matrices

is always singular (see the Remark at the close of Section 1.9). Hence we have a contradiction. and we must have k = n. Thus Rule (i) is proved. The basis e l , . . . , en defined in (1.105) is called the standard basis of V " . For n = 3 this is the familiar basis i, j, k. In general,

We now prove a theorem of much importance for the geometry of E n :

THEOREM

If u and v are vectors of V n ,then J u . vl 5 lul Ivl

(Cauchy-Schwarz inequality).

(1.107)

The equality holds precisely when u and v are linearly dependent. Furthermore, lu

+ vl i lul ;t Ivl

(triangle inequality).

(1.108)

The equals sign holds precisely when u = hv or v = h u with h 1 0.

Proof. If v # 0 and u is not a scalar multiple of v. then u and hence lu tvl' > 0 for all t , so that

+

(U

+ tv) . ( U + tv) > 0

or

+

+ tv # 0 for every scalar t,

P ( t ) = (u12 2 t ( u . V)

+ t71vlZ> 0

for all t . Hence the discriminant of the quadratic function P ( t ) must be negative; that is, so that lu . vl < lul Ivl. Similarly, l u . vl < lul Ivl if u # 0 and v is not a scalar multiple of u. If u = cv (in particular, if u = O), then

by RuleX in (1.102) and the definition (1.101). Similarly, equality holds ifv = c u (in particular, if v = 0). Thus (1.107) is proved, along with the conditions for equality. To prove (1. log), we write:

The last inequality follows from (1.107). By taking square roots we obtain (1.108). The investigation of the case of equality is left to Problem 8 following Section 1.15. __i)

Now let PI,P2.Pj be three points of E n . By (1.951, ( 1 . 9 7 e d (1.1011, 1 PIP21 equals the distance d from PI to P2. We write d ( P 1 ,P2) = 1 PIP21.It follows that d ( p t , P ~L ) Oand that d(P1.P2) = 0 only for PI = P'. Also, d ( P I , P 7 )= d ( P Z , P I ) , d __i) d since I P2 PII = l - PIP7 l = I PIP2l. Finally, from ( l .97), PIP3 = P I PZ P2P3, SO that, by (1.108),

-

+

49

50

Advanced Calculus, Fifth Edition that is, the length of one side of a triangle is less than or equal to the sum of the lengths of the other two sides. This explains the term triangle inequality in (1.108). Because of the theorem, we can define the angle 8 between two nonzero vectors u, v by the conditions

for by (1.107), u . v/(lul Ivl) lies between -1 and 1 (inclusive). With the aid of distance and angle, one can now develop geometry in n-dimensional space in familiar fashion. We define two vectors u and v to be orthogonal (or perpendicular) if u - v = 0, that is, by (1.110) if they form an angle of n/2 (or if one of the vectors is 0). We define an orthogonal system of vectors as a system of k nonzero vectors v l , . . . , vk such that each two vectors of the system are orthogonal; the set of vectors is necessarily linearly independent (Problem 9 following Section 1.15), so that by Rule (g) above, k 5 n. An orthogonal system of unit vectors is called an orthonormal system. There exists an orthonormal system of n vectors, namely, e l , . . . , en. Every orthogonal system of n vectors forms a basis for Vn,by the preceding Rule (h); we call such a system an orthogonal basis (or, for a system of unit vectors, an orthonormal basis). Given k linearly independent vectors vl , . . . , vk, it is always possible to construct an orthogonal system of k vectors ul , . . . , uk,each of which is a linear combination of v l , . . . , vk (Gram-Schmidt orthogonalization process). We illustrate this for k = 3; the construction is then easily pictured in 3-dimensional space:

To obtain u2, we subtract from v2 the vector obtained by "projecting" v2 on vl; to obtain u3, we subtract from v3 its projections on ul and u2.By (1.1 1 l), ul = vl, uz is a linear combination of vl and v2, and u3 is a linear combination of v l , v2, v3; also, vl = ul , v2 is a linear combination of ul and u2,and v3 is a linear combination of u l ,u2, and uj. The proof of these assertions and of the orthogonality is left to Problem 11 following Section 1.15. The process clearly extends by induction to any finite number of linearly independent vectors. Furthermore, in general the process is such that, for each h, uh is a linear combination of v , , . . . , vh,and vh is a linear combination of ul , . . . , uh. Remark. With the aid of the Gram-Schmidt process we can extend Rule (j) as follows:

1) If k < n and ul, . . . , uk form an orthogonal system in V n ,then there exist . . . , un such that u l , . . . , un form an orthogonal basis. vectors The proof is left as an exercise (Problem 10(j)).

Chapter 1

Vectors and Matrices

Volume in E n . In 3-dimensional space, we saw that &D = &u . v x w gives the volume of a parallelepiped whose edges, properly directed, represent u, v, w (Section 1.4). Furthermore, D equals a third-order determinant, as in Eq. ( I .33). These results generalize to E" as follows: If v l . . . . , v,, are vectors of Vn, then

equals plus or minus the volume of an n-dimensional parallelotope whose edges, properly directed, represent vl . . . . . v,,. In Eq. ( 1.1 12), vl = ( U I I , . . . , UI,,),. . . , v, = ( u n 1 , . .2nn). The parayotope can be chosen to have vertices 0 , P I , . . . Pn, where v l = OPl, . . . v,, = OPn and the n-dimensional solid consists of all points P of En such that

.

.

where0 5 t l 5 I . . . . . O 5 t,, 5 1 (see Sections 11-15 and 11-16 of CLA).

q . 1 5 AXIOMSFOR V" From Rule (h) of Section 1.14 we can state: XIII. V" has a basis consisting of 11 vectors. It is of considerable intercst that Rules I through XI1 of (1.102) plus Rule XI11 can be regarded as a set of axioms. from which one can deduce all other properties of the vectors in n-dimensional space. without reference to the sets of components. For example, u 0 = u, since

+

+

by Rules VI, VII, and IV. Also, the equation v u = w has a solution for u-namely, the vector w (- l)v, which we also denote by w - v; for

+

+

by Rules 11, VI. IV, VII, I, and the rule u 0 = u just proved. The solution is unique, forv+u=wimplies-v+(v+u)=-v+w=w-vor(-v+v)+u=w-v or u = w - v; here we wrote -v for (- l)v, as is usual. Other algebraic properties can be deduced in similar fashion. Also, the norm Iul is defined in terms of the scalar product by Eq. (1.101); its properties derive from those of the scalar product. In particular, the Cauchy-Schwarz inequality and triangle inequality follow as before, without use of components. One can introduce coordinates or components on the basis of the axioms themselves. for Rule XI11 guarantees existence of a basis v,, . . . , v,,. The Gram-Schmidt process can then be used to construct an orthogonal basis u, . . . . u,. We "normalize"

.

51

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Advanced Calculus, Fifth Edition these vectors by dividing by their norms to obtain an orthonormal basis:

For an arbitrary vector v we can now write, in unique fashion, and call (vr, . . . , v;) the components of v with respect to the basis ey, . . . , e;. We note that

because the basis is orthonormal. In general,

On the basis of Rules I through XI1 we now verify (see Problem 12 below) that

This simply means that the vector operations can be defined in terms of components just as in (1.98), (1.99), and (1.100). Accordingly, while the new basis may be different from the original one (this is simply a choice of new axes), all properties described with the old components can be found just as well with the new ones. This means that all properties of the vectors can be found from (1.102) and Rules XI11 alone. This leads to the following definition:

DEFINITION A Euclidean n-dimensional vector space is a collection of objects u, v, . . .called vectors, including a zero vector 0, for which the operations of addition, multiplication by scalars, and scalar product are defined and obey Rules I through XIII. a By virtue of our discussion there is, except for notation, really only one Euclidean n-dimensional vector space, namely, Vn . We have emphasized the axiomatic approach to vectors. There is a similar discussion for points. In fact, given a Euclidean n-dimensional vector space, we can assign components (v;, . . . , v,*)to each vector as above. To the zero vector (0, . . . , O), we then assign an origin 0 ; to each vector (vf, . . . , v,*)we assign a point P and interpret v;, . . . , v,*as the coordinates of P. All the concepts of geometry can now be introduced as above, and we recover our n-dimensional Euclidean space, the same (except for notation) as En. One can ask whether there is any value in the generalization of vectors to n-dimensional space, since we live in a 3-dimensional world. The answer is that the generalization has proved to be exceedingly valuable. The equations of a mechanical system having "N degrees of freedom" are easier to describe and understand in terms of a vector space of dimension 2N. In the kinetic theory of gases an n-dimensional space (the "phase space") is of fundamental importance; here n may be as large

Chapter 1 Vectors and Matrices as In relativity a 6dimensional space is needed. In quantum mechanics, one is in fact forced to consider the limiting case n = m; surprisingly enough, this theory of a vector space of infinite dimension turns out to be closely related to the theory of Fourier series (Chapter 7).

PROBLEMS 1. In v 4 , 1 e t u = ( 3 , 2 , 1 , 0 ) , v = (1.0, 1 . 2 ) , w = ( 5 - 4 , I , -2).

+

+

a) Find u v, u w , 2u, -3v, Ow. b) Find 3u - 2v, 2v 3w, u - w, u v - 2w. C) Find u . v, u . w, ( u J ,Ivl. d) Show that w can be expressed as a lincar combination of u and v and hence that u, v, w are linearly dependent. 2. In E n the* segment P I P2 joining points P I . PZ is formed of all points P such that t ~ 1 ~ 2 , w h e rt e5o 1. ~ a) In E4, show that (5, 10, - 1, 8) is on the line segment from (1,2,3,4)10 (7, 14, -3, 10). b) In E4, is (2, 8, I, 6) on the line segment joining ( I , 7, 2, 5) to (9,2, 0, 7)? C) In E 5 , find the midpoint of the line segment from P I : (2,0, l , 3 , 7 ) to P2: (lo, 4, -3, 3, 5); that is, find the point P on the segment such that IQI =~ q l . d) In E 5 , trisect the line segment joining (7, 1, 7, 9.6) to (16, -2, 10, 3, 0).

+

+

PIP=

-

e) In E n , show that if P is on the line segment PI P2, then I$ -.

-

I

+ IS I = I P I P2 I. __f

-2-

3. Prove the P 5 g o r e a n theorem in E n: If P I P2 is orthogonal to P2P3, then I P I P31 I P I PZl2 I P2P312.[Hint: Write the left side as

+

-

f

4. Prove thc Law of Cosines in En: If P I P2 and P I P3 are nonzero vectors, then

'+

where 8 is the angle between PI Pr and PI P3. 5. a) In E 5 , find the sides and angles of the triangle with vertices (1,2,3,4,5),

(5,4,2,3,1),

and

(2,2,2,2,2).

b) Prove: In E n , the sum of the angles of a triangle is n. [Hint: For a triangle with sides a , b, c in En, construct the triangle with the same sides in the plane E2; why is this possible? Then show, with the aid of thc Law of Cosines, that corresponding angles of the two triangles are equal.] 6. Prove the rules ( I . 102), with the aid of (1.96) through (1.101). 7. Prove Cauchy's inequality:

[Hint: Use ( 1.107).]

53

Advanced Calculus, Fifth Edition

8. Prove the rule for equality in (1.108) as stated in the theorem. [Hint: Use (1.109) to reduce the problem to that for l u . vl = lul Ivl. Use the first part of the theorem and the meaning of linear dependence to prove the desired result.] 9. Prove: The vectors of an orthogonal system are linearly independent. 10. Prove the following rules for vectors in V n ,with the aid of the hints given. a ) Rule (a) [Hint: If clvl + . . . + ckvk = 0 and, for example, ck # 0, then express vk as a linear combination of v l , . . . , vk-1. For the converse, suppose vk = clvl . . . ck-lvk-1 and write this equation in the form clvl . . . ckvk = 0.1 b) Rule (b) [Hint: Suppose vk = 0 and choose cl , . . . , ck not all 0 so that clvl . . . C k V k = 0.1 C) Rule (c) [Hint: One has clvl + . . . + ck+lvk+l = 0 with not all c's 0. Show that ck+~cannot be 0.1 d) Rule (d) e ) Rule (e) f ) Rule (f) g) Rule (h) [Hint: Apply Rules (g) and (e).] h) Rule (j) [Hint: By Rule (i), v l , . . . , vk do not form a basis. Hence one can choose vk+l so that v l , . . . , vk+l are linearly independent. If k + 1 < n, repeat the process.] i) Rule (k) [Hint: Apply Rule ( j ) to obtain a b a ~ i vl s , . . . , v , . By Rule (g), ul , . . . , uk+l, vk+l, . . . V , are linearly dependent, so that clul ' ' ' Ck+lUk+l ck+2Vk+l ' ' ' C n + l V n = 0. Show from the hypotheses that ck+2 = 0, . . . Cn+l = 0.1 j) Rule (1) [Hint: Let vl = ul , . . . , vk = uk. Then choose vk+l, . . . , v, as in Rule (j) so that v l , . . . , v, form a basis. Now apply the Gram-Schmidt process to v l , . . . , v,.] 11. a) Let vl , v2, vg be linearly independent. Prove that the vectors (1.11 1) are nonzero and form an orthogonal system. Prove also that, for h = 1 , 2 , 3 , uh is a linear combination of v l , . . . , vh, and vh is a linear combination of u l , . . . , uh. b) Cany out the process (1.1 11) in v3with vl = i, vz = i + j + k, v3 = 2j k, and graph. 12. Prove that if u ; , . . . , uz and v ; , . . . , v,* are the components of u and v with respect to the orthonormal basis e f , . . . , e i , then (1.1 16) holds. 13. (Change of basis) Let e;, . . . , e,T be an orthonormal basis of V n ,so that (1.1 14) and (1.115) hold for an arbitrary vector v . In particular, we can write

+ +

+ +

.

+ +

+

.

+ +

+ +

+

a i j = e i . e r , i = 1 , . . . , n. j = l ,..., n. a ) Show that, with A = (aij), col(v;, . . . , u,*) = A col(ul, . . . , u,). Thus the matrix A provides the link between the components of v with respect to the two orthonormal bases. [Hint: Write v = ulel +. . . + v,e, = v;e; + . . . vie; and dot both sides with ey for i = 1, . . . , n.] b) Show that the jth column of A gives the components of ej with respect to the basis e f ,. . . , ez and the ith row of A gives the components of e: with respect to the basis e l , ..., en. c) Show that A is an orthogonal matrix (Section 1.13).

+

Chapter 1 Vectors and Matrices 14. (Change of coordinates in E n ) We choose a new origin 0' and a new orthonormal basis e;, . . . , e,t for Vn as in Problem 13. Then each point P obtains new coordinates by the equation

*

O * P = x;e;

+ . . . + x,fei.

a) Let 0 * have old coordinates (hl . . . . , h,). Show with the aid of the results of Problem 13 that

.

col(x;, . . . x;) = A col(xI - h l , . . . . x,,

-

h,,),

where A is an orthogonal matrix. b) Show that if the origin is changed but there is no change of basis (so that one has a translation of axes), then

x:=x,-hi.

i = 1 , . . . , n.

A basic tool of the calculus is the concept of function. We can describe this in general terms as the assignment to each object in one set (the domain of the function) of an object in a second set (the target set of the function). The range of the function is the set of objects in the target set that are actually used (see Fig. 1.15). A function is also called a mapping. We speak of the function as a mapping of the domain into the target set. When the range is the whole target set, we replace "into" by "onto." We now consider the simultaneous linear equations

or, in matrix form,

where x = col (.rl, . . . , .rn),y = col ( y , , . . . , y m ) . (In this section, all vectors will be written as column vectors.) Through (1.1 17) or (1.1 17') a vector y of Vm is assigned to each vector x of Vn. Thus we have a function or mapping whose domain is V n and whose range is contained in V m . Furthermore, if xl and x? are in Vn and c l , c:! are scalars, then

Figure 1.15

Function or mapping.

55

56

Advanced Calculus, Fifth Edition

+

~2x2the that is, the mapping (1.117') assigns to each linear combination clxl corresponding linear combination of the values assigned to xl and x2. We call such a mapping linear: DEFINITION Let T be a mapping of V" into Vm.Then T is linear if for every choice of xl and x2 in Vn and every pair of scalars c l , c2, one has

Thus for every m x n matrix A, the equation y = Ax defines a linear mapping T of Vn into Vm with T(x) = Ax. Conversely, if T is a linear mapping of Vn into Vm, then there is an m x n matrix A such that T(x) = Ax for all x in Vn.To prove this assertion, we first note that (I. 119) implies the general rule

Now let el, . . . , en be the standard basis in Vn and let

For each vector x (1.120) that

'

c .

= (xl, . . . , x,) = xlel + . . . + x,e, in Vn, it then follows from

= A col(xl, . . . ,x,) = AX,

where A is the m x n matrix whose column vectors are ul, . . . , u,. It follows that the study of linear mappings of Vn into Vm is equivalent to the study of mappings of the form y = Ax, where A is an m x n matrix. For each linear mapping T we call A the matrix of T, or the matrix representing T. For each matrix A we call the corresponding linear mapping T the linear mapping T determined by A; we also write, more concisely, the linear mapping y = Ax, or even, the linear mapping A. From the proof just given we see that for each linear mapping T the column vectors u l , . . . , u, of the matrix A representing T are the vectors T(el), . . . , T(e,,) of Vm.Thus if A = (aij), then

Furthermore, as in the proof, we can write Thus the linear mapping A assigns to each vector x = col (xl, . . . , x,) the linear x,u, of the column vectors combination x l u l . .

+ +

U,

, . . . , U,

(vectors of Vm).

It is very helpful to think of a linear mapping as suggested in Fig. 1.16, in which the vectors are represented by directed segments from the origin; here n = 2, m = 3.

Chapter 1 Vectors and Matrices

Figure 1.16

Linear mapping of V' into V'.

Figure 1.17

Linear mapping of V2 into V'

[: :I.

57

Let the linear mapping T have the matrix A = Evaluate T ( x ) f o r x = ( 1 , O ) = x ~ , x = ( 0 ,1 ) = ~ 2 , ~ = ( 2 , - 1 ) = x ~ . x = ( 1 , - 1 ) = ~ ~ a n d graph. EXAMPLE 1

Solution. T ( x l ) = A col ( I , 0) = col(2, I), T(x2) = A col(0, 1) = col(3, 2), T(x3) = A col(2, -1) = c01(1,0), T(x4) = A col(1, -1) = c01(-1, -1) (see Fig. 1.17). We note that x l = e l and x2 = e2, SO that

the first column vector of A. Similarly, T(x2) = U? = c01(3,2), the second column vector of A . Also x3 = 2el - e2, so that Similarly, T(x4) = T(e, - e2) = ul EXAMPLE 2

u2 = (- 1, - 1).

Find a linear mapping T of T1

Solution.

-

0 = (2, 1 2) and

T has the matrix A =

[' :]

v2 into v3 such that T((0, 1)) = (5, 3,7).

2 7

Notation. We shall write T ( x l , . . . , x,) for T((x,, . . . , x,)), since the extra parentheses are not necessary.

58

Advanced Calculus, Fifth Edition For each particular linear mapping T of Vn into Vm,one is concerned with such properties as the following:

a) the range of T-the set of all y for which T ( x ) = y for at least one x; b) whether T maps Vn onto Vm-that is, whether the range of T is all of Vm; C) whether T is one-to-one-that is, whether, for every y, T ( x ) = y has at most one solution x; d) the set of all x for which T ( x ) = 0; this set is called the kernel of T ; e ) the set of all x for which T ( x ) = yo, where yo is a given element of Vm. We remark that T ( 0 ) must be 0 . For T ( 0 ) = T(O0) = OT(0) = 0 . Hence the kernel of T always contains the zero vector of Vn. With regard to (c) and (e) we have a useful rule: THEOREM Let T be a linear mapping of Vn into Vm and let T ( x o ) = yo for a particular xo and yo. Then all solutions x of the equation T ( x ) = yo are given by

where z is an arbitrary vector in the kernel of T . Hence T is one-to-one precisely when the kernel of T consists of 0 alone. Proof: We are given that T ( x o )= yo. If also T ( x ) = yo, then

T ( x - xo) = T ( x ) - T ( x o )= yo - YO = 0 . Hence x - xo is an element z of the kernel of T :

Conversely, if x = xo

+ z, where z is in the kernel of T , then

The last sentence of the theorem follows from the previous result, since T is one-to-one precisely when the equation T ( x ) = yo has exactly one solution, for every yo for which there is a solution.

EXAMPLE 3

Let T map

v2into V b n d have the matrix A=

[;%I.

Then A x = 0 is equivalent to

This is satisfied for xl = -2x2, that is, by all vectors t ( 2 , - 1 ) of v2.We note that A col(2, 3) = ( 1 6 , 24,8); that is, T ( x ) = (16,24, 8 ) is satisfied for x = xo = (2,3).

Chapter I

Vectors and Matrices

Hence all vectors x such that T ( x ) = ( 1 6,24, 8) are given by

where t is arbitrary.

EXAMPLE 4

T maps

v 3into V 7 and

To find the kernel, we must solve the equations .rl - 12 = 0, .r2 - x 3 = 0. The solutions are given by all ( x l ..r2,x3) such that .rl = .TI = .r3. hence by all vectors t ( 1 , 1 , 1 ), where t is an arbitrary scalar. Here T is not one-to-one. However, the range of T is all of v 2 ,for the equations

can always be solved for . X I , s2.xi-for example, by taking .v3 = 0 . xl_= y2, and = vl v l . Therefore T maps V 7 onto V' but is not one-to-one.

.XI

+

EXAMPLE 5 T maps V n into Vn' and T ( x ) = 0 forevery x. Thus T has the matrix 0 = Om,. We call T the zero inupping and sometimes also denote this mapping by 0 . T is clearly linear and is neither onc-to-onc nor onto. EXAMPLE 6 T maps V " into V" and T ( x ) = x for every x. Thus T has the matrix I = I,,. We call T the identipmupping and sometimes also denote this mapping by I . T is clearly linear and is one-to-one and onto. EXAMPLE 7

Let T map

v3into ~ % n dhave the matrix

Find the range of T Solution. The range of T consists of all y = .rlu l are the column vectors of A :

+ .r2u2+ ~

3 ~ where 3 .

u l , u:, uI

Thus the range consists of all linear combinations of u l ,u 2 ,u,. If u l ,u,, u3 are linearly independent, then they form a basis for v3.and the range is v 3 .However, we verify that 2ul - u2 - u 3 = 0, so that these vectors are linearly dependent. Furthermore, we see that u , , u: are lincarly independent and that u3 is expressible as a linear combination of u , . u:. Hence the range is given by all linear combinations of u l and u:, and this set is not all of v'. Thus T does not map V' onto v', and the equation T ( x ) = yo has no solution for x, for some choices of yo; for example, T ( x ) = ( 1,O. 0 ) has no solution, as one can verify (the vector ( l , O , 0 ) is not a linear combination of u I and u?).

59

60

Advanced Calculus, Fifth Edition

PROBLEMS In these problems, all vectors are written as column vectors. Also, the following matrices are

1. Let the linear mapping T have the matrix A . a ) Evaluate T(1.0). T(0, l), T(2, -I), T(-1, 1) and graph. b) Find the kernel of T , determine whether T is one-to-one. and find all x such that T(x) = (2, 3). c) Find the range of T and determine whether T maps v 2onto v 2 . 2. Let the linear mapping T have the matrix B. a ) Evaluate T(I.O), T(0, I), T(1, -I), T(-1, -1)andgraph b) Find the kernel of T , determine whether T is one-to-one, and find all x such that T(x) = (2,4). C) Find the range of T and determine whether T maps v 2onto v 2 3. Let T map V n into V m and have the matrix F. a) Find n and m . b) Find the kernel of T , and determine whether T is one-to-one. C) Find the range of T and determine whether T maps V n onto V m 4. (a), (b), (c) Proceed as in Problem 3 with matrix G. 5. (a), (b), (c) Proceed as in Problem 3 with matrix H.

6. 7. 8. 9.

(a), (b), (c) Proceed as in Problem 3 with matrix J. (a), (b). (c) Proceed as in Problem 3 with matrix K . (a), (b), (c) Proceed as in Problem 3 with matrix L Let the linear mapping T have the equation y = Cx. For general x, find the angle between x and T(x) = y, as vectors in v 2 ,and also compare 1x1 and IT(x)l. From these results, and y as where 0 is the origin interpret T geometrically. [Hint: Consider x as of E 2 and P and Q are points of E 2 . ] SO. Let the linear mapping T have the equation y = Dx. Regard x as G,y as as in Problem 9, and describe geometrically the relation between x and y = T(x).

3

3,

z,

Chapter 1

Vectors and Matrices

11. Interpret each of the following linear mappings geometrically, as in Problems 9 and 10:

12. If T maps V ninto V " ,then x and y = T ( x )arc vectors in the same n-dimensional vector space and can be eomparcd (see Problems 9, 10, and 11). If T ( x ) is a scalar multipIe of x, for somc nonzero x, say T ( x ) = Ax, thcn we say that x is an eigenvecror of T, associated with the eigrnvalue A. If T has the matrix A, then xis also an eigenvector of A, associated with the eigenvalue h, as in Section 1.1 1. Prove: If v l , . . . , vk are eigenvectors of the linear mapping T , associated with distinct eigenvalues A l . . . . , h k , then v l . . . . , vk are linearly independent. [Hint: Let T have the matrix A. Use induction. Verify that the assertion is true for k = 1. Assume 1 by assuming clvl + . . . + that it is true for k = m < n and prove it for k = m c,+lv,+l = 0, multiplying by A on the left, and then eliminating v,+l to obtain

+

Now use the induction hypothesis to conclude that cl = 0, . . . . c, = 0 and hence also C,+l = 0.1 13. a) Let T be a linear mapping of V n into V m .Prove: If v l . . . . , vk are linearly dependent vectors of V n , then T ( v l ) ,. . . . T ( v k ) are linearly dependent vectors of v m . Is the converse true? Explain. b) Let A be a nonsingularn x n matrix. Show that if v l . . . . . vk are vectors of V n ,then they are linearly independent if and only if A v l , . . . , A v ~are linearly independent. 14. a) Let u l , . . . . un be a basis for V nand let v l , . . . , v , be a basis for V m. Let T be a linear . . . p,u, of mapping of V n into V m ,so that T assigns to each vector u = plul V n a vector v = qlv1 . . q,v, of V m .Show that

+

+. +

+

where B is the m x n matrix such that

Thus for each choice of bases in V" and V m there is a matrix B representing the linear mapping T . b) With reference to part (a), let u, = ul,el

+ . . . +u,,en

for j = 1, . . . , n ,

in terms of the standard bases in V" and V m ,respectively. Show that

C)

where V is the m x m matrix ( v , ~ ) U. is the n x n matrix ( r i i , ) and y = Ax is the representation of T in terms of standard bases in V nand V mas in Section 1.16. (Why is V nonsingular?) With reference to parts (a) and (b), let T be a linear mapping from V" to V n and let only one basis be used, so that vl = u l , . . . , v,, = u , . Show that in part (b). B is similar to A . Show further that if the basis u l , . . . , un is orthonormal, then B is orthogonally congruent to A.

61

Advanced Calculus, Fifth Edition

By a subspuce of V " we mean a collection W of vectors of V" such that

(i) W contains 0, and (ii) if u, v are in W , then so is each linear combination nu

+ bv.

Let W be a subspace and let W contain vectors besides 0. We then seek sets of linearly independent vectors in W. Each such set contains at most n vectors but may in fact always contain fewer than n . We let k be the largest integer such that W contains k linearly independent vectors, and call k the dimension of W. When W contains 0 alone, we assign the dimension 0 to W . Whcn k = 11, W necessarily coincides with V n . For then W contains linearly independent vectors u l , . . . , u,,, and by repeatedly applying (ii) we conclude that W contains each linear combination of u , . . . . , u,,. Since u l , . . . . u , form a basis for v n ,W = V " . For k = I . W consists of all scalar multiples of a single vector u , , that is, all w = s l u l .If, say. n = 3. then the corresponding points P such that = wlul trace out a line L through 0 in 3-dimensional space (Fig. 1.18a). If k = 2, then W contains two linearly independent vectors u l u7 and, by (ii), contains all the linear combinations

3

.

These linear combinations exhaust W. since any vector not expressible in this form would have to be linearly independent of u l , uz (Rule (c) of Section 1.14). If again n = 3, then the corresponding points P such that = w fill a plane in space, as in Fig. 1.1 8b. In general, we see that if W has dimension k, then W consists of all linear combinations

3

.

where u l , . . . ur are linearly independent vectors of W. We term ul. . . . , uk a basis of W . There are many choices of basis for a given W , but all havc the same number of vectors (Problem 4, which follows below). Geometrically, the subspaces of V" correspond to the point 0 (k = O), and lines, planes. and "hyperplanes" through 0 , in E", and to E" itself (k = n ) .

Figure 1.18 (a) One-dimensional subspace of V '. (b) Two-dimensional suhspace of V '.

Chapter 1 Vectors and Matrices Now let v l , . . . , vk be k vectors of Vn, not necessarily linearly independent, k 1 1. We then form the set W of all linear combinations w l v l + . . . + wkvkof these vectors. We see at once that conditions (i) and (ii) are satisfied. Hence W is a subspace. If the vectors are in fact linearly independent, then, as before, v , , . . . , vk must be a basis of W, and k is the dimension of W. If. however, they are linearly dependent, then some of the vectors v l , . . . , vk can be omitted, and we see that W has dimension k' < k . We can choose k' as the largest integer such that k' of the vectors v , , . . . vk are linearly independent. For example, we can suppose by renumbering that v l , . . . , vk8are such a linearly independent set (as large as possible) in W. Then v,, . . . , vk8form a basis for W. The practical determination of a basis for W can be carried out by Gaussian elimination as in Section 1.10. We form the k x n matrix A whose row vectors are the given vectors v,. v2. . . . , vk. Thus W consists of all linear combinations of the rows of A; we call W the row space of A . We then use the steps I (adding multiples of one row to the other rows) and I1 (interchanging two rows) of Section 1.10. Neither of these changes the row space of A. For I1 no proof is needed. For I, v l , . . . , vk are replaced by v I + c I v h ,. . . , vk+ckvh,where ch = 0. Every linear combination of these vectors is a linear combination of v l , . . . , vk and is hence in W. Also, every vector in W can be written as a linear combination of the new vectors, since vh is one of the new vectors and v, = (v, + c , vh) - C , vh for i # h . Hence the row space is unchanged. As in Section 1.10, we can apply the two steps to reduce A to row echelon form with nonzero row vectors u l , . . . , u, and (for r < k), the following rows are 0. The vectors u l , . . . , u, are linearly independent as one sees at once from the row echelon form. Thus u l , . . . , u, are a basis for the subspace W and W has dimension r , equal to the rank of A , as in Section 1.10. We conclude that the rank of a matrix A is the maximal number of linearly independent row vectors of A. From the above analysis it follows that if A is k x n , then the rank r cannot exceed n or k.

.

EXAMPLE 1 Letv, = ( 2 , 4 , - I , l),v2 = ( 4 , - 2 , 1, 1),v3 = ( 2 , 14,-4,2),and let W be the set of all linear combinations of v l , v2, v3, SO that W is a subspace of v4.We seek a basis for W and form

Gaussian elimination gives 2 0 0

4 -1 -10 3 10-3

2 1

4 -10 0 0

-1

1 3 0

0

Hence ul = ( 2 , 4 , - 1, l), u2 = (0, - 10,3, - 1 ) form a basis, and W has dimension 2 = r =rankofA. One can also show that the rank of A equals the maximum number of linearly independent column vectors of A. If A is k x n , the column vectors are n vectors of Vk, say, v l , . . . , v,, and the corresponding linear combinations.form a subspace, which is the range of the mapping y = Ax from Vn to vk (Section 1.16). Thus our

63

64

Advanced Calculus, Fifth Edition assertion is equivalent to the statement that the rank ofA equals the dimension of the range o f A . For a proof see Problem 7 following this section. We can summarize our results as follows: column rank of A = row rank of A = dimension of range of A EXAMPLE 2 We again consider the matrix A of Example 1. The range of A consists of all linear combinations of wl = col(2.4.2).

W?

= col(4. -2. 14),

We find a basis for the range by Gaussian elimination applied to the matrix (transpose of A):

We obtain

Therefore z l = co1(2,4. 2), z2 = col(0, -10, 10) are a basis for the range; the dimension of the range is 2 = r , as expected. The kernel. The kernel of A consists of all x in V n such that Ax = 0. The kernel satisfies (i) and (ii) and is hence a subspace of V". For A 0 = 0 and if Au = 0, Av = 0, then A ( u u bv) = a A u + bAv = 0. We let h be the dimension of the kernel of A ; h is called the nullity of A.

+

Relation between rank and nullity. h . Then we have

Let A be a k x n matrix of rank r and nullity

To show this, we consider the corresponding homogeneous simultaneous linear equations

As we did in Section 1.10, we can solve these by Gaussian elimination, reducing A to the row echelon form. The solutions are given by linear expressions for r of

Chapter 1

Vectors and Matrices

the unknowns in terms of the remaining unknowns. By renumbering we can assume that x l , . . . , x, are expressed in terms of sr+l.. . . , x,,:

Here we write

Then the solutions are given by

where w , + ~.,. . , w , are arbitrary scalars. Thus the kernel of A is represented as the set of linear combinations of n - r vectors u , + ~.,. . u,,.We see that these vectors are linearly independent (see Problem 5 below). Therefore the kernel has dimension n - r or h = n - r . Equation (1.124) is proved. [Note: As in Section 1.10, if r = 0, Ax = 0 is satisfied by all x, so that h = n.]

.

Maximum rank. Determinant deJinition o f rank. For a k x n matrix A we have seen that the rank r of A cannot exceed k or n. When r is the largest integer satisfying this condition, A is said to have tnaximurn rank. If k 5 n, this means that r = k; if k > n, it means that r = n . By a minor of A we mean a determinant formed from the array A by striking out certain rows and columns to obtain a square array. If, for example, A is 3 x 3 , then A has one 3 x 3 minor det A , nine 2 x 2 minors, and nine 1 x I minors (the nine entries of A ) . One can show that r , the rank of A , is the largest integer such that some r x r minor of A is nonzero (see Problem 6, which follows).

EXAMPLE 3 We again consider the matrix A of Example 1. The kernel of A is found from the Gaussian elimination applied to A , as was done in Example 1. Hence it consists of all x such that

- 10-x2+ 3x3 - X', = 0. We solve for x , , x2, letting x3 = w 3 ,x4 = w 4 .We find

65

66

Advanced Calculus, Fifth Edition Thus

x = ( x l , X?, X1. .r4) =

where -oc < w3 <

00, -oc

(-

3

w3 - -w4. 10

< w4 <

3 K W

-

1 10

-W4.

Wj,

w4

oc.Accordingly,

are a basis for the kernel. T h u s h = 2 = 4 - r as in E q . (1.124). We c a n also check that the 3 x 3 minors o f A are all 0 : 4

-1

but there are nonzero 2 x 2 minors:

H e n c e again w e find r = 2.

PROBLEMS 1. Find the rank of the matrix in two different ways and verify equality:

+

2. (a), (b), (c), (d) Find the nullity h of each matrix of Problem 1 and check that h r = n ; also find a basis for the kernel if h > 0. 3. Let W be the subspace formed of all linear combinations of the vectors given. Find a basis for W: a) ( 1 , 3 , 2 ) 3(2, - 1 , 2 ) , (1, 10,4), (0, 1, I),

b) ( 3 , 6 , 2 ) , ( 1 . 3 . l ) , ( 5 , 1,7),

c) ~ 1 . 2 , 2 , 3 ~ , ~ 3 . 2 , 2 , 1 ~ , ~ 1 , 1 , 1 , 1 ~ . ~ 2 . 1 . 1 , 0 ~ . d) (3, 1, 5 , 0 , I), (2. 0 , 4 , 2 . 0 ) , ( 1 , -1, 3 . 4 , -11, (1. l.O,O,O). 4. Let W be a subspace of V n of dimension k . Prove: Every basis of W has k vectors. [Hint: Show first that no basis can have more than k vectors. Next suppose that a basis has fewer than k vectors. By completing the given basis of k vectors to a basis for V n ,show that one could then obtain a basis for V n of less than n vectors.]

5. Show that the vectors u , , ~ . . . . , u, in Eq. (1.126) are linearly independent.

1

Chapter 1

Vectors and Matrices

6. Prove that the evaluation of rank by nonzero minors is correct. [Hint: Show that this rank is unaffected by Gaussian elimination and then find its value for a matrix in row echelon form.] 7. Prove: the row rank of A equals the column rank. [Hint: Use the result of Problem 6 and Rulc I of Section 1.4 to show that A and A' have thc same row rank.] 8. For a given A and given n x n matrix A , let W hc the set of vectors x such that Ax = Ax. Show that W is a subspace of V n . (If A is an eigenvalue of A. W consists of 0 plus all eigenvectors for this 1.) 9. Prove: For a given y, the equation Ax = y has a solution if and only if the matrices A and [ A y] have the same rank. Here [ A y] is obtained from A by adjoining the column vector y at the right. [Hint: Consider the relationship between y and the range of A as a subspace.]

'

Thus far in this chapter we have considered only Euclidean n-dimensional vector spaces. One is often led to consider sets of objects that can be added and multiplied by scalars, in accordance with the familiar rules, but for which there may be no scalar product or norm. We call such a set. with the given operations, a vector space. The term "Euclidean" is reserved for a vector space having a scalar product for which all the rules (1.102) of Section 1 .I4 are satisfied. In each vector space, one can define linear independence and dependence in the usual way. One then calls the vector space n-dimensional if it contains n but no more than n linearly independent vectors. For some vector spaces, one can find n linearly independent vectors for every positive integer n ; one then speaks of an infinite-dimensional vector space. For technical reasons it is also useful to introduce a zero-dimensional vector space V o , consisting of 0 alone; this vector space has no sets of linearly independent vectors. For convenience we define a vector space formally: Definition. A vector space V is a collection of objects u, v, . . . called vectors, including a zero vector 0, for which addition and multiplication by real scalars are defined in accordance with the following rules: IV. ( a + b)u = au + bu. VT. lu = u. VII. ou = 0

By a subspare of V we mean a subset W of V satisfying the two conditions (i) and (ii) of Section 1.17. We observe that the rules (1.127) must also hold in such a subspace W . Thus W is also a vector space. For many vector spaces the objects called vectors do not arise in a geometrical context, and our standard boldface or arrow notations are not appropriate; see, for instance, Examples 1 and 2 below. As was pointed out in Section 1.15, other rules can be deduced from Rules I through VII-for example, the rule stating that subtraction is always possible and is unique and the rule u 0 = u. Furthermore, the discussion of basis and linear independence carries over to an arbitrary vector space. In particular, Rules (a), (b),

+

68

Advanced Calculus. Fifth Edition (c), (d), and (e) of Section 1.14 remain valid. Rules (f) and (g) are essentially the definition of an n-dimensional vector space. Rules (h), (i), (j), and (k) remain valid in the following forms:

h') If v l ,. . . , v, are linearly independent vectors in an n-dimensional vector space V, then vl,. . . , v,,form a basis for V. i') Every basis for an n-dimensional vector space consists of n linearly independent vectors. j') If k < n and v l ,. . . vk are linearly independent vectors in an n-dimensional vector space V, then there exist vkfl.. . . , V, SO that vl,. . . , v,,form a basis for V. k') If vl, . . . . vk are linearly independent vectors in a vector space, V and u l ,. . . , u k + areall ~ expressibleas linearcombinations o f v I ,. . . , vk,then u , , . . . , uk+~ are linearly dependent. The discussion of a general n-dimensional vector space V can in fact be referred back to the case of V n by the following procedure. Let vl.. . . , vn be n linearly independent vectors in V. Since V does not contain n 1 linearly independent vectors, it follows that every vector of V is expressible uniquely as a linear combination of vl.. . . . v,,,so that vl.. . . , v, form a basis. Hence for every vector u in V we can write

+

and, as in Section 1.14, we call u I ,. . . , u, the components of u with respect to the basis vl.. . . , v,,.We can identify the vectors of V with the corresponding n-tuples, writing u = (u . . . , u,). Then

exactly as in V". Hence (apart from considerations related to the scalar product or norm), V is the same as Vn. For a general, not necessarily finite-dimensional, vector space V, one can take advantage of the theory of V n in the following way. Let v,,. . . , v, be linearly independent vectors in V. Then all the linear combinations of vl, . . . vn themselves form a vector space V', as in the preceding paragraph, and we can identify V' with V n by replacing each vector of V' by the corresponding n-tuple of components with respect to v,,. . . , v,,.Hence V' forms an n-dimensional vector space. We call V' an n-dimensional s~rbspaceof V . As a special case of the preceding paragraph, we conclude that if all vectors in a vector space V are expressible as linear combinations of n linearly independent vectors v l ,. . . , v,,,then V is n-dimensional and v l ,. . . , v, form a basis for V. We proceed to give some examples of vector spaces:

.

+ +

EXAMPLE 1 Let V consist of all polynomials in x : 3 - x , 5 x 2x2, 7 3x2 9 x 3 , . . . . The sum of two polynomials is again a polynomial; a scalar times a polynomial is again a polynomial:

+

Chapter 1 Vectors and Matrices There is a zero polynomial: 0. Rules I through VII are satisfied, as one sees at once. For example, if p(x) and q(x) are polynomials, then p(x) q(x) = q(x) p(x) (equality meaning identity). Thus Rule I holds. Hence V is a vector space. We observe that the n polynomials 1 . x , .r2,. . . . .rn-' are linearly independent. Suppose that

+

+

+

+

+

that is, let c l c2x . . . cnxn-I coincide with the 0 polynomial and hence have the value 0 for all x. Now a polynomial of degree k has at most k roots, and hence c1, . . . , c, must all be 0. Therefore I , x , . . . , xn-I are linearly independent. Accordingly, V is an infinite-ditnensional vector space. • EXAMPLE 2 Lct V consist of all polynomials of degree at most 3, that is, of all polynomials of the form a. a l x azx2 a3x3.The sum of two such polynomials is again a polynomial of degree at most 3; there is a similar statement for a scalar times such a polynomial. Rules I through VII all hold. Therefore V is a vector space. The four polynomials I , .r, x', x3 are in V and are linearly independent. Thus we can consider V as the space V', as before, formed of all linear combinations of four linearly independent vectors in the vector space of Example 1. Accordingly, as before, V is a 4-dimensional subspace of the vector space of all polynomials. a

+

+

+

EXAMPLE 3 Let V consist of all 2 x 2 matrices. Then again addition and multiplication by scalars yield matrices of the same size and, by the rules of Section 1.7, all of Rules I through VII are satisfied. The "zero vector" is 0 = OZ2.In V the four matrices

are linearly independent. For the equation

is equivalent to

[t:

0 ::]=o=[0

0 01

and hence to c l = 0, . . . , c4 = 0. Furthermore? every 2 x 2 matrix A is expressible uniquely as a linear combination of E l 1 , . . . , E22: A = (a;,) = all Ell . . . a22E22.It follows, as before, that V is 4-dimensional and that E l l , . . . , E22 form a basis for V. By similar reasoning we show that for each fixed choice of m and n the set V of all m x n matrices forms a vector space of dimension mn. The mn matrices Eij, having 1 as the ij-entry and all other entries 0, form a basis for V. •

+

+

Comp1e.x vector spaces. One can extend the concept of vector space by allowing the scalars to be complex numbers. One then obtains a complex vector space. The previous theory carries over with the obvious changes.

69

70

Advanced Calculus, Fifth Edition

EXAMPLE 4

T h e set o f all complex numbers forms a complex vector space, w h o s e dimension is 1, with a basis consisting of a n y o n e nonzero complex number.

EXAMPLE 5

T h e set of all polynomials with complex coefficients forms an infinite-dimensional complex vector space.

+

T h e set o f all functions of f o r m ae2'-' be-"" , where a and b a r e complex constants, forms a complex vector space of dimension 2, with basis

EXAMPLE 6 e2i.t

p-2i.r

PROBLEMS 1. Show that each of the following sets of objects, with the usual operations of addition and

multiplication by scalars, forms a vector space. Give the dimension in each case and, if the dimension is finite, give a basis. a) All polynomials of degree at most 2. b) All polynomials containing no term of odd degree: 3 + 5x2 + x4, x 2 - xI0, . . . . c) All trigonometric polynomials:

+ a1 cos x + bl sin x + . . . + a, cos nx + b, sin n x . d) All functions of the form a e r + be-'. a"

e ) All 3 x 3 diagonal matrices. f ) All 4 x 4 symmetric matrices A ; that is, all matrices A such that A = A'. g ) All functions y = f (x), -ca < x < ca,such that y" + = 0. h) All functions y = f (x). -CX< x < ca,such that y"' - q' = 0. i) All functions f (x) which are defined and continuous for 0 5 x 5 1. j) All functions f ( x ) which are defined and have a continuous derivative for 0 5 x 5 1. k) All infinite sequences: X I . .r,. . . . , x,, . . . . I) All convergent sequences. 2. Show that the four polynomials 1, I + x . 1 + x + x'. 1 + x + x' form a basis for the vector spacc of Example 2. 3. Show that thc following matrices form a basis for the vector spacc of Example 3:

4. Show that the functions cos2x, sin2x form a basis for the complex vector space of Examplc 6.

Suggested References Birkhoff, G., and S. MacLane, Survey of Modern Algebra. 4th ed. New York: Macmillan, 1977. Cullen, Charles G., Matrices and Linear Transformation, 2nd ed. Reading, Mass.: AddisonWesley, 1972. Curtis, Charles W., Linear Algebra: An Introductory Approach. New York: Springer-Verlag, 1999.

Chapter 1 Vectors and Matrices Gantmachcr, F. R., The Theory of'Matrices, trans. by K . A. Hirsch, 2 vols. New York: Chclsea Publishing Co., 1960. Gerald, Curtis F., and Wheatlcy, Patrick O., Applied Numerical Analysis. 6th cd. Reading, Mass.: Addison-Wesley, 1999. Kaplan, Wilfred, and Donald J. Lcwis, Calculus and Linear Algebra, 2 vols. New York: John Wiley and Sons, Inc., 1970-1971. Kuiper, Nicolaas H., Linenr Algebra and Geometry trans. by A. van der Sluis. New York: Interscience, 1962. Perlis, Sam, Theory ($Matrices, New York: Dover, 1991.

71

Differential Calculus of Functions of Several Variables

If to each point (x, y ) of a certain part of the xy-plane is assigned a real number i, then z is said to be given as a function of the two real variables x and y. Thus

are such functions. Many such functions are considered, without special mention, in the theory of functions of one variable. For example, the function y = ax is a function of both a and x , as is the function y = log, .r. The basic theorems on differentiation relate to the functions y = u v. y = u . v , y = ulv, that is, to certain functions of u and v, where u and v are functions of x . In many cases a function of one variable can be considered as a function of two variables with one of the two variables held fixed; for example, y = x3 is obtained from y = x" (a function of x and n ) by assigning to n the value 3. Similar remarks apply to functions of three or more variables. Thus

+

give u as a function of three and four variables, respectively. It will be seen throughout the following discussion that the theory of functions of three or more variables differs only slightly from that of functions of two variables. For this reason, most of the emphasis will be on functions of two variables. On the other hand, there are basic differences between the calculus of functions of one variable and that of functions of two variables.

74

Advanced Calculus, Fifth Edition Functions of two, three. four, and even millions of variables occur in physics. The following are simple examples: RT p =

L==

--

v

(ideal gas law).

xr4~i1

2 m 2

(torque on a wire).

N

(a;

+

u;

+

,=I

energy of an ideal gas in terms of velocity

w;

) (components of the N molecules

2.2 DOMAINS AND REGIONS Most of the theory of functions of one variable is given in terms of a function defined over an interval: a 5 x 5 b. For functions of .r and y a corresponding concept is needed. A natural one would be a rectangle: a 5 .r 5 b , c 5 v 5 d . But many problems require more complicated sets, such as circles, ellipses, and so on. In order to have sufficient generality to cover all practical cases, it is necessary to formulate the concept of a domain. The general term set qf points in the .xy-plane means any sort of collection of points. finite or infinite in number: the points (0,0) and ( l , O ) ; the points on the line y = x; the points inside the circle .r' + y' = I . A neighborhood of a point (.rI, will mean the set of points inside a circle having center (.rl, y l ) and radius 6; we can thus speak of a neighborhood of radius 6. Each point ( x , v) of the neighborhood satisfies the inequality:

A set of points is called open if every point (.rl, y , ) of the set has a neighborhood lying wholly within the set. The interior of a circle is open, as is the interior of an ellipse or a square; these open sets are defined by inequalities such as the following: 7

.

+

< I;

,

.ry- + - < 1; 3 2

1.~1 <

I and IyI

I.

i

The entire .xy-plane is open, as is a half-plane such as the "right half-plane": x > 0. However, the interior of a circle plus the circumference is not open, for no neighborhood of a point on the circumference lies entirely in the set. A set E is called closed if the points of the plane that are not in E form an open y' = 1 form a closed set. The set. Thus the points on and outside the circle x' points of the circumference themselves form a closed set, as do the points on and interior to the circumference. A set is called bounded if the whole set can be enclosed in a circle of sufficiently large radius. Thus the points of the square, 1.r I 5 1 , ( v I 5 1 , form a bounded set; this 2y' < 1, form a bounded set is also closed. The points interior to an ellipse, .T' open set.

+

+

Chapter 2 Differential Calculus of Functions of Several Variables A nonempty open set is called a connected open set or a domain if, besides being open, it has the property that any two points P , Q of the set can be joined by a broken line lying wholly within the set. Thus the interior of a circle is a domain. (As in Section 1.16, the word domain is also used for the set on which a given function is defined. The context will make clear in which sense the word is to be understood. Furthermore, in many cases the domain of definition of a function is also a domain as defined here.) We remark that a domain D cannot be formed of trvo nonoverlapping open sets. For example, the points for which 1x1 > 0 form an open set E composed of two parts: the set of points for which x > 0 and the set for which x < 0. The set E is not a domain, since the points (- 1,O) and (1, 0) lie in E but cannot be joined by a broken line in E (see Problem 8 after Section 2.4). A boundap point of a set is a point every neighborhood of which contains at least one point in the set and at least one point not in the set. Thus the boundary points of the circular domain x 2 y' < 1 are the points of the circumference: x 2 Y 2 = 1. No boundary point of an open set can belong to the set; however, every boundary point of a closed set belongs to the set. An inferior point of a set is a point having a neighborhood that is contained in the set. Thus every point of an open set E is an interior point of E. A boundary point of a set cannot be an interior point. The term region will be used to describe a set consisting of a domain plus, perhaps, some or all of its boundary points. Thus a region may be a domain (if no boundary points are included). If all boundary points are included, the region is called a closed region; it then necessarily forms a closed set. (See Problem 13 following Section 2.4). Thus a circle plus interior, x' y' 5 1, is a closed region. A domain is sometimes called an open region. It will be found that for most practical problems a domain is defined by one or more inequalities, and the boundary of a domain is defined by one or more equations, whereas a closed region is given by a combination of the two; for example,

+

+

+

.ry < I is a domain, .xy = 1 is its boundary, .TV

5 1 is a closed region.

These concepts are illustrated in Fig. 2.1. The extension of these ideas to three or more dimensions is not difficult; for four or more dimensions, graphical representation is essentially hopeless. Thus a neighborhood of a point (xl , y I , z l ) in space is the set of points (x, y , z ) inside a sphere,

and the other definitions can be repeated without change. In general, in n-dimensional space E n (Section 1.14) a neighborhood of radius 6 of a point A: (al , . . . , a,) consists of all points P : ( x , , . . . , x,) such that d(A, P ) < 6, that is, such that

75

76

Advanced Calculus, Fifth Edition Boundary

Figure 2.1

Set concepts.

The other concepts, such as open set and closed set, are defined in terms of neighborhoods as before. The definitions can also be adapted to the case of one dimension. A neighborhood of a point xl of the x axis is an interval: xl - 6 < x < xl 6. A dornain on the x axis is one of the following four types of sets: (1) an open interval: a < .r < b; (2) an infinite open interval: a < x; (3) an infiniteopen interval: x < b; (4) the entire x axis. A bounded closed region on the x axis is a closed interval: a 5 x 5 b. Occasionally it is convenient to use the term neighborhood of a point to mean an open set containing the point.

+

.

2.3 FUNCTIONAL NOTATION LEVELCURVES AND LEVELSURFACES Most of the functions to be considered will be defined in a domain or, occasionally, in a closed region. The notation "z = f (x, y) in domain D" will mean that z is given as a function of x and y for all points in a domain D of the xy-plane. The variables x and y are called independent variables, while z is dependent. Similarly, one writes: "u = f (x, y, z). in domain D" or "w = f (x, y , z, u) in domain D" for functions of more than two variables. As with functions of one variable, the functional notation also serves to indicate corresponding values for a given function. in the domain Thus if z = f (.r. y) is defined by the equation z = .x2+y' < l,then f(O.O)= 1, f ( j , ; ) = & , a n d s o o n . The functional relationship z = f (x, y) is sometimes written thus: z = z(x, y). For functions of three or more variables one writes, similarly, u = u(x, y, z), w = W(X,y. Z. u), y = y(xl . . . , ~ n ) . A function of two variables can be represented graphically by a surface in 3-dimensional space, as shown in Fig. 2.2. For functions of three or more variables this representation is not available. Another method for representing functions of two variables is that of level curves or contour lines. This is the method used in making a contour map or topographical map. One plots the loci,

Jw

,

Chapter 2 Differential Calculus of Functions of Several Variables

Figure 2.2

Function of two variables.

Figure 2.3

Level curves

for various choices of the constants c , , c 2 .. . . ; each locus f (x, y) = c is called a level curve of f (x. y); it may actually be formed of several distinct curves. This is illustrated in Fig. 2.3, in which f (x, y) = x 2 y x 2 2y2; the value of c is shown on each curve. The level curves often provide a better understanding of the function than a sketch of the surface z = f (x , y ). The method just described is available in principle for functions of three variables; here one draws the level surfaces: f (x, y , z) = cl , f (x, y , z ) = c2, . . . for appropriate choices of c , . c 2 ,. . . . The surfaces of constant gravitational potential (approximately spheres) about the earth illustrate this. The surfaces of constant temperature or pressure are of importance in meteorology. One can also represent a function of three variables with the aid of level curves; if, for example, z is fixed, then f (x, y, z) becomes a function of x and y and can be represented by its level curves in an xy-plane as earlier. If this is done for several values of z, one obtains a corresponding number of diagrams that together represent the function. This is common practice in meteorology, in which lines of constant pressure (isobars) are plotted for various altitudes. For functions of four or more variables the loci, f = const, are "hypersurfaces" in a space of four or more dimensions. These level hypersurfaces are mainly of theoretical interest. To represent the function graphically, one is forced to fix one or

+ +

77

78

Advanced Calculus, Fifth Edition more of the variables and thereby obtain level surfaces in 3-dimensional space or level curves in a plane. A function f ( x , y ) is said to be bounded when ( x , y ) is restricted to a set E, if there is a number M such that 1 f ( x , y ) / < M when ( x , y ) is in E. For example, z = x 2 y 2 is bounded, with M = 2 , if 1x1 < 1 and ( y 1 < 1. The function z = tan ( x y ) is not bounded for J x y 1 < in.

+ +

+

2.4 LIMITS AND CONTINUITY

.

Let z = f ( x , y ) be given in a domain D, and let ( x l y l ) be a point of D or a boundary point of D. Then the equation lim f ( x . y )

x-x,

= C

v-Vl

means the following: given any t > 0, a 6 > 0 can be found such that for every ( x , y ) in D and within the neighborhood of ( x i , y , ) of radius 6 , except possibly for ( x I , y ) itself, one has

In other terms, if ( x , y ) is in D and

then ( 2 . 3 ) holds. Thus if the variable point ( x . y ) is sufficiently close to (but not at) its limiting position ( x , , y , ) . the value of the function is as close as desired to its limiting value c . If the point ( x l , y , ) is in D and

then f ( x , y ) is said to be continuous at ( X I ,y , ) . If this holds for every point ( x , , y , ) of D, then f ( x , y ) is said to be continuous in D. The notions of limits and continuity can be extended to more complicated setsfor example, to closed regions. The preceding definitions can be repeated essentially without change. Thus if f ( x , y ) is defined in a closed region R and ( x l , y l ) is in R. then ( 2 . 2 ) is said to hold if, for given t > 0, a 6 > 0 can be found such that ( 2 . 3 ) holds whenever ( x , y ) is in R and is within distance 6 of ( x , , y l ) , but not at ( x , , y l ) . If (2.5) holds, then f ( x , y ) is continuous at ( x l , y , ) . Similar definitions hold i f f ( x , y ) is defined only on a curve in the xy-plane. The notions of limits and continuity must always be considered in relation to the set in which the function is defined. Continuity for functions of two variables is a more subtle requirement than for functions of one variable. Such a simple function as

is badly discontinuous at the origin, without becoming infinite there. For z has limit 0 if ( x , y ) approaches the origin on the line x = y , has limit 1 if ( x , y ) approaches the origin on the x axis, and has limit - 1 if ( x , y ) approaches the origin on the y axis.

Chapter 2

Differential Calculus of Functions of Several Variables

Thus no limiting value can be assigned at (0, 0). It should be noted that the level curves of this function are straight lines, all passing through (0, 0); this alone shows that there is a discontinuity at the origin. However, the fundamental theorem on limits and continuity holds without change:

THEOREM Let u = f ( x , y ) and v = g(x. y) both be defined in the domain D of the xy-plane. Let lim f'(.r, y ) = ul.

I+\\

J@,g(x, y) = U I .

v+y,

Y-?I

Then

lim

.r +x 1

f(x, Y) = . y)

11 I

(uI # 0).

-

tll

If f (.r, y) and g(x, y) are continuous at ( x i . y l ) , then so also are the functions

provided, in the last case. g ( x l , y l ) # 0. Let F(u. ti) be defined and continuous in a domain Do of the uv-plane and let F[f(.r, y ) , g ( . r . y ) ] bedefinedfor(x, y ) i n D.Then, i f ( u I , v l ) i s i n Do,

If f ( x , y) and g(.r. y ) are continuous at ( x l ,yl). then so also is F [ f (x, y), g ( x , y)]. Proof. We first consider the composite function F [ f (x, y), g(x, y)], which is the most fundamental notion of the whole theorem. Since F [ u , v] is assumed to be continuous in Do, one has

lim 'I F[u. LI]= F[irl. tll];

18

Ld

"I

L,l

(2.1 1)

by (2.6), as (.r , y) approaches ( X I y~ , ). ( u , v) approaches ( u I , v,), so that by (2.1 l),

-

lim F[f'(.r, y), g(x. y)] = F5,1@, f ( x . y), .rlim.t 1 g(x, y)] = F [ u l , vl].

X'.lI

v+

v+

Vl

v+

Yl

Vl

Thus (2.10) is established. If f and g are continuous at (xI, y,), then f ( x i , y,) = u I and g ( . r ~y, l ) = v,, so that by (2.10),

-

lim. X I F [ f ( x . y). g(x. y)] = F [ f

.I

y-r

(XI.YI).

Vl

that is, F [ f (s,y ) . g(x. y)] is continuous at

(XI,

yl).

XI. ?I)];

79

80

Advanced Calculus, Fifth Edition

+

Now one verifies easily that the particular F[u, v] r u v is continuous for all values of 14 and v. If (2.10) is applied to this choice of F , one finds lim [ f ( x . y )

.r +.r,

+ ~ ( x?.)I.

=UI

+VI.

y+y,

which is (2.7); by the same reasoning, one concludes that if f (.r, y) and g(x, y) are continuous at (.rl. v l ) , then so is f (x. y) g(.r, y). The statements about products and quotients follow in the same way by conulv. One need only show sideration of the special functions F r u . v and F that these functions are continuous (for v # 0 in the second case). This can be done directly by applying the preceding c . S definition or as follows. One shows that the functions u v and u - v are continuous functions of u and v and also that aw and w 2 are continuous functions of w (theorems on functions of one variable). It follows from the function-of-function rule just proved that (u v)' and (u - v)' are continuous and hence that

+

-

+

+

-

is continuous for all (14, v) and we obtain the conclusions about the product f . g. Next one shows that I / v is a continuous function of v for v # 0 (function of u . ( I l v ) is a continuous function of u and v one variable) and hence that u/v for v # 0. If, for example, vl > 0, then we take Do to be the half-plane v > 0 in the uv-plane. From (2.6)we conclude that g(.r, y ) > v1/ 2 > 0 in some neighborhood D l of radius 6 of (.rI,? I ) (except possibly at ( X I , y , ) itself). In the discussion of limitsand continuity, (.r. v ) can be restricted to D l ,so that the values (u, v) = ( f (x, y ), g(x, y)) lie in Do, where F = u/v is continuous. Hence the previous reasoning applies to u/v. The theorem above can be restated in analogous form for functions of three or more variables and, in the case of the composite function, for combinations of functions of one and two variables, one and three variables, and so on: t ) ( t ] . ... By virtue of this theorem, one can conclude that polynomial functions such as F [ f (-r, y)I,

F [ f ( f ) . g(t)l,

F [f (.r, ?I. :)I.

F f t

are continuous for all values of the variables, whereas rational functions such as

x'y - .r w = l-x'-y are continuous except where the denominator is 0. Mappings,from V" to Vm. At times, one considers sets of functions of n variables: ? . I=

XI.. . . ,.T,,). (2.12)

ym =

trn(ll,. . . .X,,).

Chapter 2 Differential Calculus of Functions of Several Variables We assume that all functions are defined in a domain D of n-dimensional space. Equations (2.12) then describe a mapping from D to m-dimensional space. A special case of such a mapping is a linear mapping. as discussed in Section 1.16, with D all of n-dimensional space. EXAMPLE

Here m = n = 2, and D consists of V' minus the vector (0, 0). We can use vector notation and replace f l ( x l , . . . , x,) by fl(x), and so on. In fact we can replace (2.12) by one vector equation

Here x is a vector of V", y is a vector of Vm, and f is a vector function: f = ( f l . . . . . fm). We call the mapping continuous at x0 in D if, for each c > 0, there is a S > 0 such that If(x) - f(x0)l < r for each x in D satisfying Ix - xOI < S. The previous discussion extends to such mappings. We observe also that the mapping (2.12') is continuous at x0 = (xp, . . . , .r:) if and only if all the functions f i (x, , . . . , x,), . . . , jin ( x i . . . . , x,, ) are continuous at (xp , . . . , xf ) (see Problem 10 which follows). Instead of using vector notation, we can think of ( x l , . . . , x,,) as a point in En and ( y l , . . . , y,) as a point in E m .Then we write (2.12) as

where P is a point of D and Q a point of E m .The definition of continuity then reads: F is continuous at Po in D if for each E > 0 there is a S > 0 such that, for P in D,

Here d(A. B) stands for the distance between A and B in the appropriate space: E n for d ( P , Po), E m for d ( F ( P ) , F(Po)).As in Section 1.14, in E n , for example, one has the properties: (i) d(A, B) ? 0.

d(A. B) = 0 only for A = B,

(ii) d(A. B) = d(B, A), (iii) d(A. C ) 5 d(A. B)

+ d(B. C)

These properties of distance turn out to be crucial for discussions of continuity. Because of their importance and generality, one uses the term tnetric space for any set with a real distance function d satisfying (i), (ii), and (iii). Thus En or V" can be regarded as metric spaces with distance d(A. B ) = la - bl for A = ( a l , . . . , a,). B = (b,, . . . . b,,)regarded as points or vectors.

81

82

Advanced Calculus, Fifth Edition

PROBLEMS 1. Give several examples of functions of several variables occurring in gcometry (area and volume formulas, law of cosines, and so on). 2. Represent the following functions by first sketching a surface, and second, drawing level curves: a) i = 3 - x - 3 y c ) z = sin(x

b) z=x'+y2+1

+y )

d ) z = eAY

3. Analyze the following functions by describing their level surfaccs in space:

+ y 2 + z2 uj = x 2 + y 2 z

b) u =.r+\,+z

a ) u = x2 c)

>

d ) w = x-

-

+y-

7

4. Determine the values of the following limits. wherever the limit exists: a ) lim 1-0 y+0

X:

I

-

b ) lim 7);7

\I'

+ s-,+ \.?

,+a

\--0

( I + \.')sin., c ) lim F

d ) lim ,+o

x-to y-to

.\-

+ Y-

I+.Y-)

.r-+v'

v+o

5. Show that the following functions arc discontinuous at ( 0 , 0 ) and graph the corresponding surfaces: a)z=*

b)

z

= log(x2

+ !.)'

6. Describe the sets in which the following functions arc defined:

7. Prove the theorem: Let f ( x , y ) be defined in domain D and continuous at the point ( x , 11) of D . If f ( x l y l ) > 0, then there is a neighborhood of ( X I , yl ) in which f ( x . ?;) > ( X I, y 1 ) > 0. [Hint: Use E = ( x l , y j ) in the definition of continuity.] 8. Let D be a domain in the plane. S h o v ~that D cannot consist of two open sets E l . E2 with no point in common. [Hint: Suppose the contrary and choose point P in E l and point Q in E 2 ; join these points by a broken line in D. Regard this line as a path from P to Q and l e t s be distance from P along the path, so that the path is given by continuous functions x = x ( s ) . .y = y ( s ) , 0 5 s 5 L , with .Y = 0 at P and s = L at Q. Let , f ( s ) = -1 if ( ~ ( s )y, ( s ) ) is in E l and let f ( s ) = 1 if ( ~ ( s )y,( s ) ) is in E r . Show that f ( s ) is continuous for 0 5 s 5 L. Now apply the intermediate v a l u t ~theorml: If f ( x ) is continuous for a 5 x 5 b and f ( a ) < 0, f ( b ) > 0 , then f ( x ) = 0 for some .r between a and b (see Problem 5 following Section 2 . 2 3 ) . ] 9. Prove the theorem: Let f ( x , y ) be continuous in domain D. Let f ( x , y ) be positivc for at least one point of D and negative for at least one point of D. Then f ( x , y ) = 0 for at least one point of D. [Hint: Use Problem 7 to conclude that the set A where f ( . r , J,) > 0 and the set B where f ( x . y ) < 0 are open. If f ( x , y ) # 0 in D , then D is formed of the two nonoverlapping open sets A and B; this is not possible by Problem 8.)

if

Remark I In

.

if

This result extends the intermediate value theorem to functions of two variables.

this hook. log x denotes the natural logarithm of .r

Chapter 2

d

LO. ~n v". 1x1= z:i

Differential Calculus of Functions of Several Variables

+ . . . + I:r,

as in Section 1.14.

a) Show that if 1x1 < c , thcn 1x11 < for n = 2 .

t . .. .

. Ix,, I

<

c. Interpret the result geometrically

b ) Show that if I.rl 1 < 6, . . . , lx,,I < 6 , thcn 1x1 < nS. C) In ( 2 . 1 2 ) .Ict J I( X I , . . . , .r,). . . . , frn(x1. . . . , x , , ) be defined in domain D and contlnuous at ( x y . . . . . x 4 ) . Show that the corresponding mapping ( 2 . 1 2 ' ) is continuous at xo. [Hint: Givcn c > 0, choose S > 0 so small that If ( X I , . . . .r,,) - fl (.r:, . . . x:)l < e l m , . . . . I f ; , ( x r . . . . , X I , ) - fnr ( x : , . . . , x;)l < E / I ~for 1x - xO\ < 6. Conclude from (b) that If(x) - f(xo)l < c for Ix - xol < 6.1 d ) Show that if the mapping (2.12') is continuous at xo, then each o f the functions ~ I ( - T. .I. .. .r,,). . . . . f , ( x ~ .. . . , x,,) is continuous at ( x : , . . . x:). 11. An (infinite) sequence of points P I , . . . . P,,, . . . in the plane is said to converge and have limit Po:

.

.

.

lim P,, = Po

n+h;

or

P,,

-+

Po

if for each E > 0 there is an integer N such that d ( P , ,. P o ) < c for n > N . Show that the limit Po is unique. [Hint: If P,, -, Po and P,, -+ P;, Pi # Po, then take E = ; t l ( f i ) . P i ) and obtain a contradiction.] 12. Show that a set E in the plane is closed if and only if for every convergent sequence (Problem 11) of points { P , , )in E, the limit of the sequence is in E. [Hint: Suppose E is closed and PI, -+ Po, with P, in E for all n. If Po is not in E, thcn choose c > 0 such that d ( P , Pu) < 6 implies that P is not in E (why is this possible'?) and obtain a contradiction. Next suppose E is such that whenever { P , ) is in E and P,, -+ Po, then Po is in E . To show that E is closed, suppose that Po is a point not in E and that no neighborhood of Po consists solely of points not in E. Then choose Prr such that P,, is in E and d( PI,. Po) < l l n . Show that PI, -+ Po and obtain a contradiction.]

13. a) Show that the empty set is open (and closed) and hence the whole plane is closcd. b ) Show that the set formed of a set E in the plane and its boundary is closed

Let r: = f ( . r , y) be defined in a domain D of the .ry-plane and let (.r I , yl) be a point of D. The function f (.r, y l ) then depends on .r alone and is defined in an interval about . r l . Hence its derivative with respect to x at .r = xl may exist. If it does, its value is called the partial derivative o f f (s.y) with respect to x at ( . r l . yl ), and is denoted by

One has thus, by the definition of the derivative,

af -(.rl, i)x

y l ) = lim A.r -U

+

. [ ( X I A X . ?.I)- f ( x l , Y I ) A.r

When the point ( x , . yl) at which the derivative is being evaluated is evident, one ~ a j 1 a . r for the derivative. Other notations commonly used can write simply i j z l i ) or

83

Chapter 2

Differential Calculus of Functions of Several Variables

The preceding definitions extend at once to functions of three or more variables. If w = ( x , I,. U . v ) , then one partial derivative at ( x , y , u , v ) is

aw

-=

au

lim g ( x , y . u

+ Au, V ) -g(x, y, U . V ) Au

Au+O

(2.14)

When only three variables x , y , z are involved in a discussion, the notation az/a.x is self-explanatory: .x and y are independent, and y is held constant. However, when four or more variables are involved, the partial derivative symbol by itself is ambiguous. Thus if x , p, u , v are involved, then a u / a x may be interpreted as f , ( x , y), where u = f ( x , y ) , v = ~ ( xy);, one may also interpret a u / a x as h , ( x , y , v ) , where u = h ( x , y , u ) . For this reason, when four or more variables are involved, it is advisable to supplement the partial derivative symbol by indicating the variables held constant. For example,

(g)v

means f , ( x . y ) . where r = f (x. y ) .

(ayv

means h , ( x . y , v ) , where u = h ( x , y , v).

The independent variables consist of the variable with respect to which the differentiation is being made, plus all variables appearing as subscripts.

+ u - 2 v , then

EXAMPLE 1

If w = .ruv

EXAMPLE 2

If u , v , x , y are related by the equations u=x-y,

v=x+y,

then

whereas

since u can be expressed in terms of x and v by the equation u = 2.r

from which one also obtains

-

v,

85

86

Advanced Calculus. Fifth Edition EXAMPLE 3

1f x L

+ y2

-

z2 = 1 , then

whence

In forming the preceding partial derivatives & / a x and a z / a y . changes Ax and Ay in .r and y were considered separately; we now consider the effect of changing .Y and y together. Let (.r, y ) be a fixed point of D and let (.r A x . y A y ) be a second point of D. Then the function z = f ( x , y ) changes by an amount A,: in going from ( x ,y ) to(.r A x , y Ay):

+

+

+

+

This defines Az as a function of A.r and Ay ( x and y being considered as constants). with the special property Az = 0

when

Ax = 0

and

Ay = 0.

+ .ry + .ry2,then

For example, if z = x'

Here Az is of the form AZ = a A x

+bay +c

~

+'dA.rAy + e a y 2 + f

that is, a linear function of Ax and Ay plus terms cfhigher degree. In general. the function z = f ( x , y) is said to have a total differential or to be differentiable at the point (.r, y) if, at this point, A Z = ( I A X + ~ A ~A +xE+~E .? .Ay.

(2.16)

where a and b are independent of A x , Ay and t and c2 are functions of A.r and A? such that lim

€1

A1 +O A?-0

the linear function of Ax and Ay,

= 0.

lim

Ax-+0 Av--0

€2

= 0;

84

Advanced Calculus. Fifth Edition

Figure 2.4

Partial derivatives.

are z,, f , , f l or, more explicitly, 7,(xl, y l ) , f , ( s l . y l ) . f l( r , , .I). When subscripts are used, there can be confusion with the symbol for components of a vector; hence when vectors and partial derivatives are being used rogerhe~a notation s~lchas az/a.r o r af 1a.r is preferable for partial derivatives. The function z = f (x, y) can be represented by a surface in space. The equatlon y = y l then represents a plane cutting the surface in a curve. The partial derivative az/d.r at (.rl, y l )can then be interpreted as the slope of the tangent to the curve, that is, as tan a , where a is the angle shown in Fig. 2.4. In this figure, := 5 x' - v', and the derivative is being computed at the point x = 1, v = 2. For y = 2. z = 1 r', so that the derivative along the curve is 2.r; for r = I , one finds f,(1, 2) to be 2. The partial derivative is defined similarly; one now holds r constant. equal to x i , and differentiates f (.rl. v) with respect to y. One has thus

+

+

5

This can also be interpreted as the slope of the tangent to the curve in which the plane .r = s, cuts the surface z = f (x, y). One also writes f,(.r,, y l ) , f2(.rl,y l ) for this derivative. If the point ( X I , yl) is now varied, one obtains (wherever the derivative exists) a new function of two variables: the function ,f,(.r, y). Similarly, the derivative a:/ay at a variable point ( s , y) is a function f,(.r, y). For explicit functions r = f (.r, y), evaluation of these derivatives is carried out as in ordinary calculus, for one is always differentiating a function of one variable, the other being treated a s a cmstant. For 7 7 example, if ,: = .r- - y-, then

Chapter 2 Differential Calculus of Functions of Several Variables is then termed the total differential of z at the point (.r, y ) and is denoted by dz:

If Ax a n d Ay a r e suficiently small, d: gives a close approximation to A i . More precisely, one can write

where a and b are constants; by (2.17) the percentage error in each term caused by replacing t, and t 2 by 0 can be made as small as desired by choosing A x and Ay sufficiently small. (This argument fails if a or b is 0.) In the preceding example, Az has a total differential at each point (.r, y), with

and t, =

Ax

+ Ay( 1 + 2y).

t z =.ray

+ A x Ay.

THEOREM If z = f (.r , y ) has a total differential (2.18) at the point (x , y), then f is continuous at (.r , v ) and

that is. the two partial derivatives exist at (x, y ) and have the given values.

Proof.

Set Ay = 0. Then, by (2.16) and (2.17),

az ax

-

= lim

AX-o

Az -

AX

= lim AX+()

A.r(a + t i ) AX

= lim ( a A.T+O

+el) =a.

Similarly, one shows that dz/ay = b. The continuity o f f follows from f (.r .v Ay) - f (x, y ) = Az, where by (2.16) and (2.17) Az has limit 0 as Ax Ay -+ 0.

+

+ Ax, -+

0, a

Remclrk. The theorem shows that the differential, when it exists, is unique. One can describe the differential of f ( x , y ) at a point ( X I ,y I ) as the unique linear funcrion 4 . r - . T I ) + b(y -1'1) which approximates A: = f (.r, y)- f ( x i , y i ) a s i n E q . (2.16), with A.r = x - X I , Ay = y - p i . The existence of the partial derivatives at the point ( x , y ) is not sufficient to guarantee the existence of the total differential (see Problem 7 following this section); however, their continuity near the point is sufficient for this.

FUNDAMENTAL LEMMA If z = f (.r, y ) has continuous first partial derivatives in D, then z has a differential

at every point (I, y ) of D.

87

88

Advanced Calculus. Fifth Edition Proof. Let ( x , y ) be a fixed point of D. If x alone changes, one obtains a change A z in z :

this difference can be evaluated by the Mean Value theorem for functions of one variable, for, with y held fixed, z is a function of x having a continuous derivative f x ( x , y). Thus one concludes:

where x l is between x and x

+ A x . Since f x ( x , y ) is continuous, the difference

approaches zero as A x approaches 0 . Thus

Now if both x and y change, one obtains a change A z in =f (x

z:

+ A x , y + A y ) - f ( x , y).

This can be written as the sum of terms representing the effect of a change in x alone and a subsequent change in y alone:

+

A z = [ f ( x A x , Y ) - f ( x , y)l

+ [f ( x + A x , y + A y ) - f ( x + A x . y ) ] .

(2.22)

The first term can be evaluated by (2.21). The second is evaluated similarly, with z a function of y alone:

where yl is between y and y the difference

+ A y . It follows from the continuity of f,(.r,

y ) that

approaches 0 as both A x and A y approach 0. One has nbw

Equations (2.21), (2.22), and (2.23) now give

where and c2 satisfy (2.17). Thus the Fundamental Lemma is proved.

z has a differential d z as stated in (2.20),and

Chapter 2

Differential Calculus of Functions of Several Variables

For reasons to be explained, Ax and Ay can be replaced by dx and dy in (2.20). Thus one has

which is the customary way of writing the differential. The preceding analysis extends at once to functions of three or more variables. For example, if w = f ( x , y , u , v), then

EXAMPLE 1

If: = x2 - y 2 , then dz = 2x dx - 2y d y .

EXAMPLE 2

If w = ?, then dw =

dx

+ i dy 2 dz. -

PROBLEMS 1. Evaluate

2 and * if ify

b ) z = y sinxy C ) X~ + X ~ Y -

xzz

+

z3

-2=

o

d) z = -J e) z = ( x 2

+ y2)3/2 + 2y)

f ) z = arc sin ( x

g) P"+2eY-e7 - z = O h) x y 2

+ ?z2 + xyz = 1

2. Acertain function f ( x , y ) is known to have the following values: f (0,O)= 0 , f (1.0)= 1, f ( 2 , O ) = 4, f ( O , 1 ) = -2, f ( 1 , l ) = -1, f ( 2 , l ) = 2 f(O 2 ) = -4 f ( l 2) = -3. f (2. 2 ) = 0. Compute approximately the derivatives f,(l, 1) and f."(l, 1 ) . 3. Evaluate the indicated partial derivatives: 3

a)

(%Iv and ( $ ) - r

if u = x 2

-

y Z ,u = x

-

2y

(g)"and if x = eUcos u , y elf sin u ( g ) and y (g),, if u = x - 2 y , = u 2)) d) (g)? and if r = Jw, x = r cos R b)

=

(%)u

c)

u

-

($)-r

4. Find the differentials of the following functions: a)z=$

b ) z = log JiF7 ~

)

x).

e) z = arc tan

z

5

=

~

7

7

7

89

90

Advanced Calculus, Fifth Edition 5. For the given function z = , f ( . r , ti nd A; and d ; in terms of A.r and Ay at .r = I . Comp;lrc these two functions for sclcctcd values of A x . A , near 0 . !I),

Y

= 1.

6. A certain function 2 = f'(.r,p) is known to havc the valuc , f (1 , 2) = 3 and derivatives f,( 1. 2 ) = 2. f,.( 1 . 2 ) = 5. Make "reasonablc" estimates of , f ( I . I , 1.8). f ( I .2, 1.8), and . f ( 1.3, 1.8).

7. Lct

x)

+

;= ,f!.r. = .ry/(.r2 y 2 ) except at ( 0 ,0 ) :let f ' ( 0 . 0) = 0. Show that ilz/i).r and i ) : j i ) y exist for all (.r. J ) and arc continuous except at ( 0 , 0 ) .Show by thc Fundamental

Lemma that; has a differential for (.r. y ) # ( 0 .0)but not at (0.0).since f is discontinuous at (0. 0). IIt is instructive to graph thc level curves of f .]

For a function of n variables

the differential is obtained as in Section 2.6:

Thus it is a linear function of d.rl. . . . . d.r,,, whose coefficients f',,. . . . , f,,,are the partial derivatives of f at the point considered. This linear function is a close approximation to the increment Ay in the sense described in Section 2.6:

t l -+

0. . . . . 6 , + 0

as

cl.rl

+ 0 . . . . . dx,, -+ 0.

On occasion, one has to deal with several functions of n variables:

If these functions have continuous partial derivatives in a domain D of E n . then all have differentials:

Chapter 2 Differential Calculus of Functions of Several Variables These equations can be written in matrix form:

Thus the vector col ( d y , , . . . , d y m )is obtained from the vector col (dxr , . . . , dx,,) by multiplication by the ma1ri.x

This matrix is called the Jacobian matri.~of the set of functions (2.29); its entries are partial derivatives of the functions (2.29), evaluated at a chosen point of D. Equations (2.29) assign a point ( y l . . . . , y,) in E m to each point ( X I ,. . . , x,) of D. Thus they describe a mapping of D into E"' (see Fig. 2.5, in which n = 3, m = 2). The linear equations (2.30) or (2.31) describe a linear mapping (Section 1.16) that approximates the given mapping near a chosen point; the linear mapping is expressed in terms of coordinates ( d x l , . . . , dx,,), with origin at the chosen point in D and axes parallel to the given axes, and coordinates (dy,, . . . dy,) related similarly to the corresponding point in E m . We can simplify further by regarding (2.29) as a vector function of the vector

.

X

= ( X I , . . . , X, 1:

Figure 2.5 mapping.

The differential as a linear mapping approximating a given

91

92

Advanced Calculus. Fifth Edition Here y = (I.,, . . . , y,,), and f is a ver,torfunction ( f1 , . . . , f,,). We then write (2.31 ) in the concise form:

d y = f, dx.

(2.31')

Here d x = col ( d x , , . . . , d r,), d y = col ( d v , . . . . , d y , ) , and f, 1s an abbreviation for the Jacobian matrix (af,/a,x,). We can also write a y , / a x , for i)f;/a.r,and are then led to write (2.3 1 ' ) in the form

d y = y, d x . which is much like the formula dy = yld.r for functions of one variable.

EXAMPLE 1

Hence

The function f is defined by the equations

['I] [

+

[

2x1d x i 2.rr d.r2 - 2x3 d.r3 2.r 2x1 d x - 2x2 d x 2 2 . ~ d q ] = 2.rl d y = dy2 = d ~ 3 -2.~1 d.rl + 2x2 d.r2 + 2.1.~ dx3 -2x1 At the point

( X I . .rz, .x3)

+

:;]

2.r2 -2.r3 -2.r2 2.rz

dxl kx2]. d.r3

= ( 2 , 1 , 1) we find ( y l ,y2. v,) = ( 4 , 4 . -2) and

If x = (2.01, 1.03, 1.02), then d x = (0.01.0.03,0.02). and the last equation gives

so that, approximately, y = (4.06,4.02, -1.94); the exact value is (4.0606.4.0196. - 1.9388).

EXAMPLE 2 u = x 2 - xy, tr = .x.v + y2. Here the independent variable vector is ( x , y ) , and the dependent variable vector is (14. 21). We have

At ( x , v ) = ( 2 , 1). ( u , V ) = ( 2 . 3 ) . and the approximating lincar mapping is

Chapter 2

Figure 2.6

Differential Calculus of Functions of Several Variables

Mapping and approximating linear mapping for Example 2.

(see Fig. 2.6). We study this linear mapping in more detail. For d y = 0 we have du = 3d.y and d v = d.r. so that (du, d v ) follows a line of slope {; for d.x = 0 we have du = -2dy and d l ] = 4 d y . so that (du, d v ) follows a line of slope -2. Similarly, the linear mapping can be studied along other lines. In particular, we verify that the points of the square

correspond to the points of the shaded parallelogram in the (du, d v ) diagram. The area of the square is 1, and the area of the parallelogram is l(3i + j) x (-2i

+ 4j)l =

;1

= 14 sq. units.

+

For the given nonlinear mapping 14 = .Y' - xy. v = .ry y' the lines y = 1 and ~ 0, respectively) correspond to parabolas through .Y= 2 (on which dy = 0 and d . = (2, 3) in the uv-plane, as shown. The square 2 5 A- 5 3. 1 5 y 5 2 (the same square as above) corresponds to a "curved parallelogram" as in the figure. Thus we see in a geometric way how the linear mapping approximates the given mapping. We observe that the linear mapping takes the line d y = 0 or the line d x = 0 to a line tangent to the curve obtained from the nonlinear mapping. We have seen that for the square 0 5 d x 5 1.0 5 dy 5 1 , the correA sponding region for the linear mapping is a parallelogram of area 14 = similar calculation applies to an arbitrary square, and we find that the area is always multiplied by 14. Since arbitrary figures can be approximated by unions of squares, one concludes by a limiting process that all areas are multiplied by 14 under the linear mapping.

Remark.

1: -:I.

93

94

Advanced Calculus, Fifth Edition EXAMPLE 3

The mapping y = cos u sin tl.

= cos u cos L!.

:= sin 14

has the Jacobian matrix - sin u cos v -

1

cos u sin v cos u cos L] .

-

sin u sin v cos u

0

EXAMPLE 4 Let w = F(x, y, z). Then the Jacobian matrix of F is the row vector (a F l a x , a F l a y , a Flaz). We call this vector the gradient vector of F and denote it by V F or grad F . This is discussed in Section 2.13. Similarly, F ( x l , . . . , x,,) has as Jacobian matrix the row vector (F,, , . . . , FA,!),called the gradient vector of F. We return to the general mapping y = f(x) and assume m = n, so that the mapping is given by equations

1:

Yn

= fn(x13 . . . xn). 7

In this case the Jacobian matrix y, = (afi/axj) is .square, and we can form its determinant:

We call this determinant the Jacobian determinant (or simply, the Jacobian) of the mapping. For the corresponding linear mapping d y = y, dx, J is the determinant of the matrix y, of the mapping; this determinant measures the ratio of n-dimensional volumes. Symbolically, This is justified, as in the preceding Example 2, on the basis of the determinant formula for volume of an n-dimensional parallelotope (Eq. (1.107) in Section 1.14). For n = 2, I J I measures the ratio of areas, as in Example 2 (see Problems 5 and 6 below). Since the linear mapping approximates the nonlinear one, we can say that the absolute value of the Jacobian determinant J measures the ratio of corresponding volumes for small regions near the chosen point. This relationship is studied further in Sections 2.9,4.6, and 5.14. An illustration is provided by the preceding Example 2, in which n = 2 and J = 14 at the point; this is precisely the ratio of the area of the parallelogram to the area of the square to which the parallelogram corresponds. For n = 1, J becomes the derivative d y l d x , and its absolute value does measure the

Chapter 2 Differential Calculus of Functions of Several Variables

-

ratio of corresponding lengths Ay , Ax: Idy 1d.r 1 / Ay I / I Ax 1 for small Ax, since d?-/dx = lim(Ay/A.r) as A.r -+ 0. The Jacobian determinant is also denoted as follows:

The concept of Jacobian determinant and these notations can also be applied to n functions of more than n variables. One simply forms the indicated partial derivatives, holding all other variables constant. For example, for f ( u , u , w), ~ ( 1 4 u, , w), one has

PROBLEMS 1. Obtain the Jacobian matrix for each of the following mappings: a) y~ = 5x1

+ 2x1. yz = 2x1 + 3 . ~ ~ .

+ x : , y? = 3x1x2.

b) vl = 2x:

2. Obtain the linear mapping d y = f, d x approximating the given mapping y = f(x) near the specified point and use the linear mapping to obtain an approximation to the value f(x) specified. a)

7

=x i

+ x;,

1

b) vl = xlxz

-

= ~ 1 xat2 (2, 1). approximate f(2.04, 1.01).

+

x:. y: = ~ 1 x 2 x l x i at (3, 2, l), approximate f(3.01, 1.99, 1.03).

C) 11 = ex cosy. u = e.'sin y, uj = 2e' at (0. n/2),approximate value of (u.,u, u ) ) for (x. v ) = (0. I . 1.6).

. . . . . yn =x:+...+x:-l

d) ?I =xl+...+.r:.y2 =xf+x;+...+x: approximate f( I . 0.1, . . . .O. 1).

-

3. Obtain the Jacobian determinant requested: a) ,)(.,.,y ) for u = .r3- 3xV2,u = 3x2? - y3 b, C)

~ I ( u , v, U I )

act, g,

for 11 = .re?cos z. u = xe? sin I, w = x e '

for ,f(u. u, Ul) = u'uw, g ( u

ti,

w ) = u2u2 - u,4

a t ( 1 , o, . . . , 01,

95

96

Advanced Calculus, Fifth Edition

4. For the mapping ir = e' cosy. = P' sin y from the .ry-plane to the irv-planc. carry out the following steps: a ) Evaluate the Jacobian determinant at ( 1, 0). I )

h) Show that the square R,,: 0.9 5 x 5 1.1. -0.1 5 y 5 0.1 corresponds to the region R,,,, bounded by arcs of the circles u' + u2 = 0 ' ', u 2 + 11' = r2.' and thc rays I ! = fitan 0.1 ) ( 1 . u L 0, and find thc ratio of the area of R,,,. to that of' R,,. Compare with the result of (a). c) Obtain the approximating linear mapping at ( I . 0) and find the rcgion R,:, corrcsponding to the square R , , of part (b) undcr this linear mapping. Find the ratio of thc arca of R:,,, to that of R , , and compare with the results of parts ( a ) and tb). 5. a) Let u, v be linearly independent vcctors in v'. Show by gcomctric rcasoning that the points P of the plane for which

fill a parallelogram whosc edges, properly dircctcd. rcprcscnt u and v. b) With u, v as in (a). Ict A be a nonsingular 2 x 2 matrix. so that A u . A v arc also linearly indcpcndcnt (Prohlcm 13 following Section I. 16)and under the linear mapping y = A x the parallelograrn of part (a) is mapped onto a parallclogram givcn by

in the plane. Show that thc arca of the second parallelogram is ldct A ( times thc arca of the first. [Hint: Lct B hc thc matrix whose column vectors arc u . v and let C bc the matrix whosc column vectors arc A u . A v . Show that the areas in question arc ldct B ( and ldct CI.] 6. Gcncralizc the results of Problein 5 to 3-dimensional spacc. Thus in (a) use threc vcctors u . v. wand consider the corresponding parallclcpipcd and in ( h ) take A to he a nonsingular 3 x 3 matrix and considcr volumcs.

2.8 DERIVATIVES AND DIFFERENTIALS OF COMPOSITE FIJNCTIONS T h e functions to be considered in the following will be assumed t o be defined in appropriate domains and t o have continuous first partial derivatives, so that the corresponding differentials can b e formed.

THEOREM

IS: = f'f.r.

y) and .r = ~ ( 1 ) y. = h ( t ) . then

IS := J'(.I-, y ) and .r = girr. v), y = h ( u , v). then

Chapter 2

Differential Calculus of Functions of Several Variables

In general, if z = f ' ( x . y , r, . . .) and .r = g ( u , v . w . . . .), y = h ( u , v . w . . . .), r = p ( u , v. w , . . .). . . . , then

These rules, known as "chain rules," are basic for computation of derivatives of composite functions. Equations (2.33). (2.34), and (2.35) are concise statements of the relations between the derivatives involved. Thus in (2.33).

is the function of r whose derivative is denoted by d ; / d r , while d x l d r and d y l d t stand for g'(r) and h'(r ), respectively. The derivatives a,-/ax and a z / a y , which could be written (ilz/a.r), and ( a z j a y ) , , stand for f,(.r, y ) and f,(.r, y ) . In (2.34),

is the function whose derivative with respect to u is denoted by B:/au, which should be understood as ( a z l a u ) , .A more precise statement of the first equation in (2.34) would be as follows:

and similar remarks apply to the other equations. The proof of (2.33) will be given as a sample: the other rules are proved in the same way. Let r be a fixed value and let .r. y . :be the corresponding values of the functions g . h , and f . Then, for given A t , A.r and A y are determined as A . =g t

+A

-g

while A z is then determined as

By the Fundamental Lemma, one has

Hence

A v = h(r + Ar) - h ( r ) ,

97

98

Advanced Calculus, Fifth Edition As A t approaches 0, A x / A t and A y l A t approach the derivatives d.r/dt and d y l d t , respectively. while r l and 6 2 approach 0, since A.1- and A v approach 0. Hence

that is,

as was to be proved. The three functions of t considered here-x h ([)]-have differentials

= g(t). y = h(t). := f [ g ( t ) .

From (2.33). one concludes that

that is. that

But this is the same as (2.24). in whichdl and d y are A.Yand A?, arbitrary increments of independent variables. Thus (2.24) holds whether x and y are independent and d z is the corresponding differential or whether x and y , and hence z . depend on t . so that d.r, d v , d z are the differentials of these variables in terms o f t . Similar reasoning applies to (2.34). Here u and v are the independent variables on which x , v , and z depend. The corresponding differentials are

But (2.34) gives dz =

a z av ax arn)A. (-- a.r +-(-a.r au a y au a.rav ayav1 +

+

a2

A]!

Again (2.24) holds. Generalization of this to (2.35) permits one to conclude: THEOREM

The differential formula

99

Chapter 2 Differential Calculus of Functions of Several Variables

.

which holds when ,- = f ( x . y , t . . . .) and d.r = A.r, d y = A y , dt = A t , . . . remains true when x , y , t , . . . , and hence z , are all functions of other independent variables and d x , d v , d t , . . . d: are the corresponding differentials. •

.

As a consequence of this theorem, one can conclude: Anv equation in di'erenrials that is correct for one choice of independent and dependent variables remains truefor any other choice. Another way of saying this is that any equation in differentials treats all variables on an equal basis. Thus if

at a given point, then

is the corresponding differential of x in terms of v and z . An important practical application of the theorem is that in order to compute partial derivatives, one can first compute differentials, pretending that all variables are functions of a hypothetical single variable (for example, t ) , so that all the rules of ordinap dlferential calculus apply. From the resulting differential formula. one can at once obtain all partial derivatives desired. EXAMPLE 1

If z =

, then

by the quotient rule. Hence

EXAMPLE 2

If r' = x'

(g),

= ,:

EXAMPLE 3

and hence

-

y'. then r dr = .r d.r

>(:,

=

.:

+ y d y . whence

( )

If z = arc tan y1.r (x # O), then

?.

=

r -

and so on.

a

100

Advanced Calculus, Fifth Edition

PROBLEMS 1. If (a) y = u + v , ( b ) y = u . v , (c) y = u l v , where u and v are functions of x , then apply (2.33) to find d y l d x . 2. If y = u u , where u and v are functions of x , then find d v l d x by (2.33). [Hint: (a')' = a x log a , (xu)' = axU-' .I 3. If y = log, v , where u and v are functions of x , then find d v l d x by (2.33). [Hint: (log,,x)'= l / [ x l o g a ] , l o g , a = l / l o g u x . ] 4. If z = r' cosy, while x and y are implicit functions of t defined by the equations

then find d z l d t for t = 0. [Note that x = 0 and y = 0 for t = 0.1 5. Let z = x 3 - 3 x Z y ,where x and y arc functions of t such that f o r t = 5, x = 7 . 1 = 2 , d x l d t = 3, and d y l d t = -1. Find d z l d t for t = 3. 6. Let z = f ( x , y), where f x ( 4 , 4 ) = 7 , f y ( 4 , 4 ) = 9 , x = 2e3' t2 - t 2, y = 5e3' 3t - 1. Find d z l d t for t = 0.

+

+

+

[Hint: Use the chain rules to evaluate the derivatives on the right-hand side.]

8. If w = f ( x , y ) and x = u cosh v , y = u sinh v , then show that

[Cf. hint for Problem 7.1

9. If z = f ( a x

+ b y ) , show that

10. Find a z / a x and a z / a y by first obtaining d z : a ) z = log sin ( x - 1) ~ ~ ~

Jm

b) z = x Z y 2

c ) x 2 + 2 y 2 - z2 = 1

11. If f ( x . y ) satisfies the identity

for a fixed n , f is called homogeneous of degree n . Show that one then has the relation

This is Euler's theorem on homogeneous functions. [Hint: Differentiate both sides of the identity with respect to t and then set t = 1.] 12. (The Stokes total time derivative in hydrodynamics) Let w = F ( x . y. z , t ) , whcrc x = f ( t ) , y = g ( t ) , z = h ( t ) , so that w can be expressed in terms of / alone.

Chapter 2

Differential Calculus of Functions of Several Variables

Show that

Here both dwldr and awli31 = F,(x, y , z. t ) have meaning and are in general unequal. In hydrodynamics, d x l d t , d y l d t , d z l d t are the velocity components of a moving fluid particle, and dwldr describes the variation of w "following the motion of the fluid." It is customary, following Stokes, to write DwlDr for d w l d t . [See H. Lamb, Hyclrodynnmics, 6th ed. (Dover: New York, 1945). p. 3.1

On occasion, o n e deals with two sets o f functions:

and

If one substitutes the functions (2.38) in the functions (2.37), o n e obtains composite functions

Under the appropriate hypotheses, o n e can obtain the partial derivatives of these composite functions by chain rules, a s in the previous section:

ayi ax,

avi a ~ ,

ayi au, a ~ ax., ,

- - -- +...+---

aul ax.,

( i = 1 . . . . , rn, j = l . . . . . n ) .

(2.40)

T h e formulas (2.40) can be expressed concisely in matrix language. T h e partial derivatives a y i / a x i are the entries in the tn x n matrix

101

102

Advanced Calculus, Fifth Edition This is the Jacobian matrix of the mapping (2.39) (see Section 2.7). The formulas (2.40) involve two other Jacobian matrices:

The chain rules (2.40) state that the product of the last two Jacobian matrices equals the previous one:

This equation is called the grrzrral chain rlrle. It includes all the rules of the previous section. EXAMPLE1 Let.vl = l r l u z - u l u 3 , y - , = u l u 3 + n i , . 111 = . r l c o s . r ~ + ( . r I - . r 2 ) ' . u2 = .rl sin.r2 .rl.r?,u3 = .rf - .rl.r? + .r$. Then, by (2.43),

+

+

143

2112

Ul

cos.rl 2(x1 - .r2) sin .r, .rl 2x1 - xz

+

-.vl

sin .t-? - 2(.rl - .r2) .r 1 cos .r2 + .r 3.Q - .r1

On the right-hand side, u 1 , ~ r 143 z . can be expressed in terms of .rl. .r?, and the two matrices can be multiplied. However, for many purposes it is sufficient to leave the result in indicated form. In particular, to obtain numerical values, one can substitute the appropriate values and multiply the matrices only as a last step. For example, for .rl = I . .r2 = 0. we obtain u l = 2, u2 = 0, u3 = I and hence

that is to say, dyl/i).rl = -7. 8yl/a.r2 = 8, a.~,/il.r~= 7, and av,/a.r2 = -4. Differentials arld the chnin rule. we obtain the equations

If we take differentials in Eqs. (2.37) and (2.38).

Chapter 2 Differential Calculus of Functions of Sevcral Variables and

au

au

du - 2 dxl+...+-'l-dx,, . - ax, ax,, In (2.44),du1, . . . , d u , are arbitrary increments A u , , . . . , Arc,,, whereas in (2.45) they are functions of the arbitrary increments d x l ( = A x l ) , . . . , d.r,(= Ax, ). However, we know from Section 2.8 that the relationships are the same no matter how we interpret the differentials. We can write these equations in matrix form:

If we eliminate the vector col ( d u . . . , d u , ) in these equations, we obtain

and so on. From these equations we can read off a y l / a x l . a y , / a . r z ,. . . . Clearly, the results are the same as (2.40)or (2.43).Thus (2.47)can be termed the general chain rule in differential form. The preceding development can be carried out even more concisely in terms of the notations of Section 2.7: The given functions are really vector functions

If we take differentials, we obtain

and hence

so that

This last equation is the same as (2.43);the previous one is the same as (2.47). We saw in Section 2.7 that a Jacobian matrix such as y, is the matrix of the linear transformation approximating the given, in general nonlinear, mapping y = f ( x ) .

103

104

Advanced Calculus, Fifth Edition Thus Eq. (2.43) asserts that if we have successive mappings u = g(x). y = f(u) and hence obtain a composite mapping y = f(g(x)),then the matrix of the linear approximation of the composite mapping is obtained by multiplying the approximating matrices of the two stages. In the special case when f and g are linear, then we have u =Bx, y = Au for appropriate matrices A and B (u, = B, y, = A), and the composite mapping is y = A(Bx) = A Bx, as in Section 1.8; thus y, = A B = y,u,. Case of square rnatrice.~. In the preceding analysis, let m = n = p, so that all the Jacobian matrices appearing are square and each has a determinant-the Jacobian determinant of the corresponding mapping, as in Section 2.7. For example,

det y. =

I

To the equation (2.49) we can apply the rule det A B = det A det B (Eq. (1.60) in Section 1.9) to obtain the following very useful rule: det y, = det y, det u,:

(2.50)

that is.

If, for example, n = 2, then each determinant here can be interpreted as in Section 2.7 as plus or minus the ratio of small corresponding areas, and (2.5 1) states roughly that AA, AA, AA, - -

--

AAx

AA,, AAx

'

where we have written AA, for an "area element" in the .rlxl-plane, and similarly for AA ,, AA,, . There is a similar interpretation for n = 3, in terms of volumes, and for higher n in terms of higher-dimensional volume.

PROBLEMS 1. Find the Jacobian matrix (ay,/ilxj) in the form of a product of two matrices and evaluate the matrix for the given values o f x l . x?. . . . . = u I u 1 - 3 u I , v ? = U: 2uIn2 + 2ti1 - u ? ; U I = X I cos3x2.u? = .rl sin3x2; a) .r 1 = 0, x2 = 0. b) I! = uy u; - 3u1 u 3 , y1 = u: - U,2 2111 - 3u3; ul = xlxlx:. U ? = . T I X ; X ~ , U3 = xfx2x3; X I = I , X? = 1, X j = 1. C) y1 = ~ d ~ e "~2' , = ulePLt2, y3 = u i ; u l = .r: + .rl, u2 = 2xf - xl; .rl = I,x2 = 0.

+

+

+

+

d) yl = ~ ~ + . . . + ~ ~ - u ~ . y ~ = u ~ + . . . + u, . ,. ?. , Jy n- =u u~ ~ + . . . + u ~ - - u ~ u1 = xf + X I X ? . U ? = X: 2x1x2,.. . ,11,, = .rI2 nxlxz: xi = 1. xz = 0.

+

+

Chapter 2

Differential Calculus of Functions of Several Variables

+

105

2. a ) Find a(:, w ) / a ( x ,y ) for x = 1, y = 0 if z = u3 3u2v - v3 + [ I ? - v 2 , w = u3 u3 - 2 u Z :u = x cos x y , v = x sin x y + x2 - y2. b ) Find a ( x , y ) / a ( s . t ) for s = 0, r = 0 if x = (:' + w ' ) ~ / y~ ,= w(;? + U I ~ ) - I / ~ :

+

:= (.\

+r + I)-',

ul = ( 2 s - r

+ I)-'.

3. Justify thc rules, under appropriatc hypotheses: a ) If y = f ( u ) ,u = g ( v ) , v = h ( x ) , then y, = y,u,v,. J I ) i ) ( ~ W ) acu, LI) acs, t ) b 4. For certain functions f ( x , y ) , g ( x , y ) , p ( u , v ) , q ( u , v ) it is known that f ( x o . .YO) = U O . ,L!(xo. Y O ) = Uo and that .fr(xo. yo) = 2, f ~ ( x oyo) , = 3, ~ r ( x oyo) , = -1, g,(xo, yo) = 5, p l r ( ~ o~ . o =) 7 , pU(uo.U O ) = I. 9 r r ( U~O ~) = , -3, q,,(uo, vo) = 2. Let := F ( x , = p ( f ( x , y ) , g ( x . y ) ) , w = G ( x , y ) = q ( f ( x , v ) , g ( x . y ) ) and find the Jacobian matrix of :(x. y), w ( x , yj at (xo. Y O ) . 5. Let ul = xi - 3x2 2 ~ ~ xU2, , = 2x1 5x2 - 3 x I x 2 Let w = ( w l , w 2 ) be a vcctor for u = ( 3 . 3 ) . Find the Jacobian matrix function of u = ( u ~u.2 ) such that w , = [: at x = ( 2 , 1) for the composite function w [ u ( x ) ] . 6. Let u = f ( x ) and v = g(x) be differentiable mappings from a domain D in 3-dimensional space to 3-dimensional space. Lct a and b be constant scalars. Let A be a constant 3 x 3 matrix. Show: a) d(u + v ) = d u +d v b) d(au bv) = adu + bdv C) d ( A u ) = A d u d ) rl(u . V ) = u . d v + v . d u e) dcu x v ) = u x d v + d u x v

+

+

':I

+

I f F(.r, y. :) is a given function of x , y , and z , then the equation

is a relation that m a y describe o n e o r several functions v' - 1 = 0, then

+ :'

z

of x and y . T h u s if x'

+

+ y 2 5 1. Either function is said to b e implicitly + y2 + z2 - 1 = 0.

both functions being defined f o r x 2

rfejined b y the equation x 2 Similarly, a n equation

may define o n e o r m o r e implicit functions w of .r, y . i . If t w o such equations are given:

1

106

Advanced Calculus, Fifth Edition it is in general possible (at least in theory) to reduce the equations by elimination to the form

that is. to obtain two functions of two variables. In general, if m equations in n unknowns are given (m < n ) , it is possible to solve for 111 of the variables in terms of the remaining n - m variables; the number of dependent variables eqiials the number of equations. The main question to be considered here is of the following type. Suppose a particular solution of the m simultaneous equations is known, for example, a quadruple ( x l , y l , z l , w,) of values satisfying (2.54); then one seeks to determine the behavior of the m dependent variables as functions of the independent variables near the given point; for example, for (2.54), one wishes to study f (n., y ) and g(x. y) near (.rl,?I), given that f (x,, y l ) = wl and g(xl. y , ) = zl. Determining behavior of a function near a point will consist here of finding the first partial derivatives at the point; when the first derivatives are known, the total differential can be found. and hence a linear appro.rimation to the function is known. It will in fact be seen that the essential step to be taken consists of a linearization of the given simultaneous equations. This has its counterpart in the formulation of laws of physics; one seeks to describe natural phenomena by means of the simplest possible equations. These are usually linear in form and are usually valid only when the variables are restricted to a narrow range. The complete description of the phenomena in general involves simultaneous nonlinear equations, which are far more difficult to grasp. A typical example is Hooke's law for a spring: F = k ' s ; the linear dependence of force F on displacement x is valid only as a first approximation. To analyze an equation of form (2.52), we assume that z = f (x. y ) is a differentiable function that satisfies the equation, so that

We assume that this relation holds for (.r. y ) in a domain D and that the points ( x , y . ;.), for ( I , y) in D and z = f (.r, y ) , all lie in a domain in which F is differentiable. Then from (2.52) or (2.56) we obtain: F, dx

+ F,, dy + F, dz = 0.

Here is considered to be the function J'(.r. y), so that dz = f , dx (2.57) we deduce that

so that (provided that F; # 0 at the points considered)

'These are the desired expressions for the derivatives.

(2.57)

+ f , dy. From

Chapter 2 Differential Calculus of Functions of Several Variables

+ + ,-' I

+ +

EXAMPLE 1 .r' y' z' = I or .r2 y' of the preceding paragraph:

-

= 0. We imitate the procedure

These equations apply to each differentiable function satisfying the given equation-in particular to the function z = ( I - r' - y')'!'. At.r = S , y = , I we find 2 = I/&, and hence

We remark that the same results can be obtained by taking partial derivutives instead of differentials in the given equation. From (2.52), with ;considered as a function of x and y , we differentiate with respect to .r and then with respect t o y to obtain

from which (2.59) again follows. In the case of Example 1 we obtain

For the pair of equations (2.54) we assume that differentiable functions u~= f (x. y ) . z = g(x. y ) satisfy the equations and can then take differentials: F, d x G , d.r

+ F , d j ! + F, d z + F,, dul = 0. + G , d ~+, G , d : + G , dul = 0.

(2.60)

We consider these equations as simultaneous lineur equations for d ; and dui and solve by elimination or by determinants. Cramer's Rule yields

d,- =

-F,dx -G,d.r

F, d y - G , dy -

G,.

I

107

108

Advanced Calculus, Fifth Edition and hence

Thus we can read off partial derivatives. Since the determinants appearing are Jacobian determinants, we can write the derivatives in terms of Jacobians:

Here we must assume that the Jacobian a ( F , G ) / a ( z ,w ) in the denominator is different from 0 at the points considered.

EXAMPLE 2

+ y2 + z2 - ZW = 0. x 2 + y2 + 22' + zw - 8 = 0. 2x2

We observe that the equations are satisfied for x = 1, y = 1, z = 1, w = 4, and seek differentiable functions z ( x , y ) , w ( x , y ) satisfying the equations near this point. If such functions exist. then

By elimination we find

and hence

dw =

x(2z

+ w )dx

z2

+

+

2y(z w ) dy, 3z2

Chapter 2 Differential Calculus of Functions of Several Variables from which we can read off partial derivatives azlax = -x/z and so on. We could have applied (2.6 1 ) directly. For example,

We observe that the determinant in the denominator equals 6z2,and this is different from 0 near z = 1. We could also have taken partial derivatives. By differentiating with respect to ,r, we obtain

+

Elimination gives az/ax = -x/z, au,/ax = x(2z w)/z2. Taking differentials saves time, since all partial derivatives are obtained at once. • The reasoning generalizes to an arbitrary set of m equations in m

+ n unknowns,

say

We seek m differentiable functions

satisfying the equations. Assuming differentiability as before, we obtain from (2.62)

where d y l , . . . , d y , are the differentials of the functions sought. Equations (2.64) are m linear equations in the m unknowns d y l , . . . , dy,,, and we have in effect linearized our problem. If the appropriate determinant is not 0, we can solve Eqs. (2.64)

109

110

Advanced Calculus, Fifth Edition

.

for d y , , . . . d?;,,.That determinant is

The equations can be solved by elimination or by determinants. We can also use matrices here. Equations (2.64)can be written:

where F , is the Jacobian matrix of the coefficients of d y l , . . . , dyn,in (2.64);that is, it is the matrix whose determinant is given in (2.65).Similarly,

Finally, d y = col (dq.1. . . . , dy,,,),and dx = col ( d . r l .. . . , d . ~ , , If) . F, is not singular at the point considered-that is, if the Jacobian determinant (2.65) is not 0-then Fy has an inverse and we can write

F y d y = -Fx dx, dy = - ( F y

)-IF,

dx.

.

This equation gives d v , , . . . d y , and hence all partial derivatives of the unknown functions f l . . . . , f,,. We can also take partial derivatives in (2.62).By differentiation with respect to Y. j . we obtain

These are m linear equations for the m unknowns a y l / a x j ,. . . , d y m / a x j .If the determinant (2.65)is not 0, we can solve for the unknowns. Cramer's Rule gives. for example,

Chapter 2 Differential Calculus of Functions of Several Variables Similar formulas are obtained for a y 2 / a x j ,. . . , a y , , , / a . ~ In , . general, a ~ i

-

a.r,

~ ( V I . .. .

~ ( F I .. ... F,,,)

.?,-I,

.tl.

? , + I .. . . , y,,,)

~ ) ( F .I ., . , F,,,) iI(y1. . . . . J,,,)

( i = 1 . . . . , m. j = I

. . . . . n ) . (2.69)

The denominator here is the Jacobian ( 2 . 6 5 ) . The numerator is obtained from the denominator by replacing the ith column by col ( a F, / a x j , . . . , a Fm/a.r,).An illustration is given by Eqs. ( 2 . 6 1 ) ,where rn = 2 and n = 2 (two equations in four unknowns). The following are additional illustrations: one equation in t w unknowns: ~ F(.r, y) = 0. Here

one equation in three unknowns: F ( x , y , z ) = 0. Here, as in (2.59),

hvo equations in three unknowns: F ( x , y , z ) = 0, G ( . r , y, z ) = 0. Here

where

three equations in five unknowns: F(.r, y , z . U , v ) = 0, G(.r, y , z , u , ,,) = 0, H(.r, y , z , u , v ) = 0. Here

For greater precision the partial derivatives appearing should have subscripts attached. For example, in ( 2 . 7 3 ) , a.r/au should be ( a x / a u ) , . The entire preceding discussion has been based on the assumption that implicit functions are in fact defined by the equations. This assumption is not always fulfilled. For example, the equations

define no functions at all. It can be shown that if the Jacobian determinant in the denominator is not 0, then the implicit equations do define functions as before. One has the following fundamental result:

111

112

Advanced Calculus, Fifth Edition

IMPLICIT FUNCTION THEOREM For i = 1. . . . , m , let the functions F i ( y l .. . . . v,,, . X I , . . . , .r,) all be defined in a neighborhood of the point Po: (Y:. . . . , 0 yn,. .xI0 , . . . . .xi:) and have continuous first partial derivatives in this neighborhood. Let the equations

be satisfied at Po and let a ( F , . . . . , F,,,) ~ ( Y .I. ..? y r n )

# 0 at Po.

Then in an appropriate neighborhood of (xp, . . . , .xf), there is a unique set of continuous functions v = ( x . .x

i = 1 . . . . , nz,

.

such that Y,! = f i ( x : . . . . x:) for i = 1 , . . . . rn and for all i

in the neighborhood. Furthermore, the f , have continuous partial derivatives satisfying (2.69). A proof for the case rn = 1, n = 1 is given in the next section. Rrmcrrk. If the Jacobian determinant happens to be 0 at the point of interest, a different choice of dependent variables (such as X I . yz. . . . , y,) may avoid the difficulty. Thus if the Jacobian matrix formed of all m n first partial derivatives of each of F , . . . , F,,, has rank m at the point Po (Section 1.17), then one can solve for n z of the variables in terms of the remaining ones.

.

+

"2.11 PROOF OF A CASEOF THE IMPLICITFUNCTION THEOREM The discussion of implicit equations in the preceding section is purely formal; it is assumed that we can solve them, and then it gives formulas for derivatives of the solutions. It is important to know when we can solve. In practice, the typical situation is that one initially knows just one point on the graph of the solution sought and then one tries to find a continuous solution through this point. The Implicit Function Theorem ensures that if the relevant Jacobian is not zero at the point, then this is possible at least in a sufJiciently small neighborhood of the initial point. The basis of the theorem is the fact that near the point our equations can be approximated by linear equations (by taking differentials), so that, provided that the relevant matrix is nonsingular, one can always solve. We here consider only the simplest case of one equation in two unknowns, which we write as

For a discussion of the general case, we refer the reader to Chapter 9 of the book by Rudin listed at the end of the chapter. (See also Retnarks at the end of this section.)

Chapter 2 Differential Calculus of Functions of Several Variables

THEOREM Let F(.r, y) be defined in an open region D of the.rv-plane, let (.ro.yo) be in D, and let F(.ro, yo) = 0; let the partial derivatives F, , F, be continuous in D and suppose that F,(xo, yo) # 0. Then there is a function

v = f (.r) with domain 1.r - .wol < 8. 6

> 0.

(2.75)

whose graph is in D, with f (xo) = yo, and that satisfies Eq. (2.74). Furthermore, a positive number q can be chosen so that the graph of (2.75) lies in the set

and provides all solutions of (2.74) in this set. The function f is differentiable, and

Proof.

We let

wherever F, # 0, so that F., = - F,.g. Now F,(.ro, yo) # 0. Let us suppose, to be specific. that F,(.ro, yo) > 0. Then, by continuity, F,.(.r , y) > 0 in a neighborhood of (xo, 1.0). heref fore we can choose 6, q so small and positive that the closed rectangular region

is in D, and F, > 0 in E. The function g is also continuous in E and hence has absolute minimum m and maximum M in E (see Sections 2.19 and 2.23):

Wereplace 6 by asmaller number. if necessary, toensure that Irn IS < q and I M 16 < q. This is done to ensure that the graphs of the linear functions

lie in E (see Fig. 2.7). We assume 6 to be so chosen. Now, since F, > 0 in E and F(.m. y o ) = 0. F is monotone strictly increasing along the line .r = .ro in E and hence is positive for y > and negative for y < vo. Along a line

F becomes a function of .r, and

dF dv d.r - Fx $- F,.-~1.r= F, +AF,. = - F , . ~ +AF,. = F,.(A - g). Hence, if A is greater than M, the maximum of g,dF1d.r is positive along the line; thus F itself is positive for .r > xo, negative for .r < .ro. There is similar reasoning. --

113

114

Advanced Calculus, Fifth Edition

Figure 2.7

Proof of the Implicit Function Theorem.

with reversal of signs, for h < rn.Thus the sign of F is as in Fig. 2.7. Since F, > 0. F is monotone strictly increasing in v on each line x = const in E and hence must go from negative to positive. Therefore by the Intermediate Value theorem (Problem 5 following Section 2.23) F ( x , y) = 0 for exactly one y for each x. This value of y we denote by f (x). Thus

and y = f ( x ) provides all solutions of the implicit equation in E. The graph o f f is squeezed between the two lines (2.76). Hence f is continuous at 10, with f (x) -+ yo = f (xo) as x -+ xo. But the same argument applies to each point (.r, y) on the graph of f ; we can find a rectangle E centered at the point and so on, as for (.YO, yo). Therefore f is continuous for all x in the interval (x - xoI < 6. (We can also define f at the endpoints of the interval, .ro f 6, but we ignore these values in the subsequent discussion.) Finally, we seek the derivative of f at xo. Along a line y - yo = h(x - .ro) with h = g(xo. vo) t . t > 0, we have, as earlier,

+

Since g is continuous at (.ro,yo), I,g(x, v ) - g(xo, yo)I < t for (x, y) sufficiently close to (x0, yo); thus the quantity in brackets is positive, and dF/dx > 0. Therefore F itself is positive along the line, for x > xo and x sufficiently close to xo. But this means that the graph of f must be below the line, that is,

for .r > xo as earlier. Similarly,

f (x) - .f (..o) X - .To

> g(xo, yo) - t

Differential Calculus of Functions of Several Variables

Chapter 2

for .r > .yo and .r sufficiently close to .Yo. Thus

In the same way the same limit is found as .r + .I-,,. Therefore

Again the argument applies to every point (.r. y ) on the graph of f', and we conclude that at each such point

Thus the theorem is proved. R e m ~ r k s . The theorem proved here gives inore information about uniqueness than

the general theorem stated at the end of the previous section: In an appropriate neighborhood of Po: (.r,).yo), (111 points (.I-, J) satisfying the implicit equation are on the graph of the solution y = f (.r) found. The general theorem can be extended in analogous fashion. The proof gives some information about the size of the ,I--interval in which a solution y = f'(.r) can be found. Specifically, one must choose a rectangle E: 1.t - .tnl 5 6. 1y - yol 5 q i n which F', > 0 (or F,, < 0 ) and in which 6 has been restricted so that 6 K < q, where K = max g(.r, ?.)I in E. One sometimes has more information about F that permits one to give a better estimate of the size of the interval. For example. if F, > 0 in E and F is positivc for ?. = yo 71 and negative for .v = yo - q. then the solulion is defined and unique for -1. - .roI < 6. The restriction Iy 5 11. in general. is needed to ensure the uniqueness of the solution. For example. the equation

+

?.'

-

y sin .r

-

e-'x

+ e' sin .r = 0

satisfies the hypotheses of the theorem with .r,, = 0. y,, = 0, and a solution is given by y = sin.\-. However, another solution is y = r ' ; we exclude this solution by restricting to a sufficiently small rectangle E about (0, 0). The existence of a unique continuous solution ?; = j'(.r) can also be shown as follows. With E chosen as above so that F, > O i n E. F(.ro,y ) is strictly increasing in y, so that F(xo,yo q) > 0. F(.Y,,. \,, - 11) < 0. By the continuity of F. for 6 sufficiently small, F(.r, yo 11) > 0 and F(.r, yo - q) < 0 for Ix - .rol < 6. By the Intermediate Value Theorem. as above. for each .r with 1.r - .rnl < 6, F(.r, y) = 0 for a unique y = f'(.r) between yo - q and y,) 11. Given c > 0, we can choose r1 so .w,,l < 6. Thus f is continuous that rl < t and we have I,f'(.t-)- f'(.ro) < c for I . \ at ro. By applying the same reasoning to each point (.xi, yl) on the graph of f we conclude that f' is continuous for 1.r - ,roi < 6 . We can extend the proof to cover the case of G(.r, y, z) = 0 with G = 0 and G, # O at Po: (.xu.yo. Y ~ ) ) say , G; > O at Po. The proof of the preceding paragraph extends at once to this case to yield a unique continuous solution z = f (.r. y ) with

+

+

+

-

115

116

Advanced Calculus, Fifth Edition graph i n an appropriate set E: 1.r - .rol < 6. iy - vol < 6. 1: - zol < q . For the equation G(.r. yo, :) = 0 we apply the theorem for the equation F(.r, \ ' ) = 0 to show that J'(.r. yo) has a continuous derivative f ,(.r yo) and the equation

.

is satisfied at ( s o , J ' ~ ) By . applying the same reasoning to each point ( . r I . yl, f (.rl. T I )) on the graph of f , we conclude that the last equation holds at each point (.r. y ) with 1.r - .roI < 6 , 1~ - yo I < 6. Since f is continuous, the right hand side of the equation is continuous Tor 1.r - x o ) < 6. ly - yo( < 6 and therefore so also is ,f,(.r, y). Similarly ,f,, is continuous and satisfies the analogous equation with G , replacing G , . A similar reasoning applies to the single equation F ( y . . r l , . . . , .r,,) = 0. to yield a solution \. = f (.xi. . . . .r,, ). For the general case of In equations F , ( y l , . . . , q,,. .rl. . . . . .r,,) = 0, i = I , . . . , m. one can proceed by induction. The case m = 1 has been disposed of. If we have proved the theorem for n1 - 1 equations, then we prove it for rn equations by observing that the hypothesis on the Jacobian d ( F l , . . . , F,,,)/iI(y, . . . . , y,,) implies that for some i , i)F,/Ry,, # 0 at Po.For proper numbering, we can take i = 111 and then solve the mth equation for y,, (by applying the theorem for a single equation); we substitute the resulting function in the previous equations; one verifies that the new equations in yl, . . . . y, , , , . r , , . . . . .r,, still satisfy the appropriate hypotheses. Hence by the induction assumption they yield desired functions y, = , f , ( . ~ -.~. ,. .x,,) for i = 1 , . . . , ni - 1 . Finally,

.

.

PROBLEMS 1. Find (;I,-/i).r),, and (i):/ijy), by Eq. (2.7 1 ): a ) 3.r- + y - - :=3 b ) xy,- + 2.1-'z + 3.x:' = 1 c) z3 +xz+2?': - 1 = O d) r-'; + cY- + ,- - 1 = 7

7

,

2. Given that

find the following partial derivatives:

3. Find (arr/ii.r), and (au/r3?;),: a) x' - y ? + + 2"' = 1. .r' + j,' - u 7 - 132 = 2 b ) tJ1 +.ru - yu - 1 = O , e ' - . r v + y r r - 2 = O c) .r2 + .LLI - yu' + 141) = 1. .rri - 2 y v = 1

Chapter 2

Differential Calculus of Functions of Several Variables

4. Given that

a ) find ciu and dv in tcrms of tix, tiy, and rlz at the point x = I , y = 1, ;= 2, u = 3, ri = 2; b) find (au/a.u),,; and (av/ay),., at this point; C) lind approximately the values of u and u for x = 1.1, y = 1.2, z = 1.8.

+

5. If xy 2xu matrix ['fl

+ 3xv + u v - 1 = 0, 2xjl + 3jju

$:I.

I,,

-

2x11+ 2uv

+ 2 = 0 , lind the Jacobian

6. Let equations Fl (.rl. x?, x j , x4) = 0 , F 7 ( x I ,x?. x j , x4) = 0 be given. At a certain point where the equations arc satisfied, it is known that

a ) Evaluatc (ijxl lilxj),, and ( a x l/i)x4),, at the point.

b) Evaluatc ( a . u l /a.u3),, and (a.r4/ax3),, at the point. c ) Evaluate ij(xl. .ur)/a(x3, x4) and a(x3, x 4 ) / a ( x l ,x?) at the point. 7. Prove: If F ( x , y , z) = 0 , then

(g),,(g)7($)x = -I.

8. In therrnociynclrnic.~the variables p (pressure), T (temperature), U (internal energy), and V (volume) occur. For each substance these are related by two equations, so that any two of the four variables can be chosen as independent, the other two then being dependent. In addition, the second law of thermodynamics implies the relation iiU

i)p

a) m - T n + p = O , when V and T are independent. Show that this relation can be written in each of the following forms: i) 7' b) m + T m i)p - p m = iJO7' T I iJ(T u) = 0 ;,(<, ),,

,

iJU

iJV

V

d) - + T n + p z = O e)

,Ip

TiJJ iJ(pV)

ilV

f ) T O ( ~ , [ hJ ) - p-iju

-

(V. P indcp.1, ( p . T indcp.),

i J ( V .T ) a ( U , ,,I

+

j ,

(U,Vindep.),

=0

I =0

(U, P i n d e ~ . ) , ( T . U indep.).

[Hint: The relation (a) implies that if

arc the cxprcssions for rlU and tip in tcrms of rlV and d T , then a prove (b), for example, one assumes relations

-

Te

+ p = 0. To

If these are solved for d U and tip in terms of d V and d T , then one obtains expressions for n and e in tcrms of a. ,fl, y , 8. If these expressions arc substituted in the equation

117

118

Advanced Calculus, Fifth Edition - T c ~ + p= 0. one hasancquation inw. fi. I/. 8 . Sincc w = ;)T/ilV,ctc.. thc rclation c way.] of form (b) is obtained. The others are provcd in ~ h samc

rr

A pair of functions

can be regarded as a mapping from thc uv-plane to the .I-!-plane (see Section 2.7). Undcr appropriate conditions this mapping is a one-to-one correspondence between a domain D l , , in the uv-plane and a domain D,, in the .ry-plane (Fig. 2.8). Onc can then consider the ir111er.c.emapping that takes each point !.r. !) in D , , to the unique point ( u , LI) such that ( 2 . 7 7 ) holds. The inverse mapping is given by functions

However, we may have difficulty in solving (2.77)to obtain these functionsexpl~citly. We may, nevertheless, consider Eqs. (2.77), in tho form

as implicit equations for the functions (2.78). We can then seek partial derivatives as in the preceding section. With F ( . r , y . 11. t i ) = , f ( u . 11) - .r and G ( . Y .y . 1 1 . 1 1 ) = g ( r t . v ) - v we have, for example.

The Jacobian determinant in the denominator is simply the Jacobian of the mapping (2.77). As earlier, we assume that it is not zero at the points considered. This condition. together with the continuity of the first partial derivatives appearing, ensures that the Implicit Function Theorem is applicable. so that the inverse mapping (2.78) is well defined. The fact that the functions ( 2 . 7 8 ) are solutions of (2.77) means that

Figure 2.8

Mapping. inves\c mapping. and curvilinear coordinates.

Chapter 2

Figure 2.9

Differential Calculus of Functions of Several Variables

Polar coordini~tesas curvilinear coorilinate.;

(If we apply a mapping and then the inverse mapping. the composite is the identity mapping.) Hence by the rule (2.5 1 ),

Thus the J~lcobicinc!f the inverse mapping i.s thc~r~c~iprot.cil of the Jacohian of' t h ~ rnr~ypirlg. Curvilinrar coorriinnres. Wc can also regard (2.77)as equations describing a transformation from rectangular to curvilinear coordinates in the .\:\.-plane. For fixed [I, i.r, y) traces a curve with u as parameter; for fixed 1 4 . (.r. 1) traccs a curve with u as parameter. The curves mentioned serve as lincs tl = const, 11 = const in the curvilinear coordinate system (Fig. 2.8). The most common example of curvilinear coordinates is provided by polar coordinates (Fig. 2.9). Here .r = r cos 8 , y = r sin 0 . The inverse functions (2.78) express thc curvilinear coordinates in terms of rectangular coordinates. For polar coordinates the inverse functions are

for an appropriate interpreration of the inverse [angcnt. The preceding discussion extend!, to 3-dimensional space and generally to E". One considers a mapping

or. in vector language, a vector function

The mapping is assurned to be defined in a domain D in E" and the functions f, are assurned to be continuous and havc continuous first partial derivatives in D, with Tacobian J( f I , . . . . , f , , ) / i ) ( r t , , . . . . I ( , , # O at the point u" = (14:'. . . . , u! j of D.

119

120

Advanced Calculus, Fifth Edition Then there is a continuous inverse mapping U I

=gl(.r I . . . . , -r,,) . . . . .

u,,

=,?,,(XI, . . . . . t-,,)

(2.81)

defined in a neighborhood (open set) Do of xi' = f(u0), whose range is a neighborhood D l of uOcontained in D , so that f(g(x)) = x

in Do.

g(f(u)) = u

in D l

(one-to-one correspondence between Do and D l ) . The inverse mapping functions g, in (2.81) have continuous first partial derivatives in D,. The partial derivatives of

these functions can be obtained as earlier by the procedures for implicit functions. One can also reason that the differentials satisfy a matrix equation

and hence d u = (fu)-I dx This shows that the Jacnbian matrix ofthe inverse mapping is sinzpiy the inverse of' the Jacobian matri.~efthe mapping (assumed to be nonsingular). In concise notation,

so that 1 det u, =det xu

and det u, det xu = 1

(2.82)

as in (2.79). The concept of curvilinear coordinates also applies to (2.80).Common examples, for n = 3, are the two systems

which are the equations of transformation from rectangular coordinates to cylindrical and spherical coordinates, respectively (Fig. 2.10).

Figure 2.10

Cylindrical coordi~latesand spherical coordinate5

Chapter 2

Differential Calculus of Functions of Several Variables

PROBLEMS I. For the tranqformation x = r. cost9, y = r. sin H from rectangular to polar coordinates, verity the relations: a ) t1.r = cos H tlr - r sin I ) (10. rly = sin H d r b ) tlr = cos fi tI I sin t9 tly . tlH = tlx

+

c)

(2)1,cos H .

d)

m=

(@) 01-

=

14(.!,

1 ~ 0H -I ,

1 )

1

= sec H ,

+ r cos H tlH + tlv

(2) = cos H . 0.r v

( g ) o= sec H

- I

rn - F

2. Given the n~apping

a ) write the equation%of the invcrsc mapping, b ) evaluate the Jacobian of'the mapping and that of the inverse mapping 3. Given the mapping

a ) compute its Jacobian, b ) evaluate

( E ) Yand ( g ) v .

4. Given the mapping -w =

with Jacobian J =

w.

f ( u , v).

?' = g ( u . U ) .

show that for the inverse functions one has

Use these results to check Problein I ( b )

5. Given thc mapping

with Jacobian

J

ill4

- ilu

-

1 il(y, 7) J i l ( u , ul)'

I ij(\,,7) J ij(u1. 1 4 ) '

- - --

ijx

,,,,.show that for the inverse functions one has

,"(-r. ?'. ') ( , A , ,,,

a~

a~

1 a(:, X ) J il(u, u : ) '

-

ilv

ilu il v

I a ( ~XI. J i J ( I L ~ .11) '

- -

-

- --

- - --

ilz

a~ il:

-

1 a ( ~ y. ) J X u . ul)'

--

I ax. Jil(u1.u)' !I)

--A

6. For the transformation (2.84) from rcctangiilar to spherical coordinates, (a) cotnpilte the ~ ~ "(" @.'., ~;#) ) , ( b ) evaluate ~ bapli).. iij + / i 3 : , ~ijHli1.r for ~ thc invcrse transfortnation (cf. Problem 5 ) .

121

122

Advanced Calculus. Fifth Edition

7. Provc that i f r = J'(rr.

1 ) ) . !,

= ,g(rr. 1 ' ) . thcn

and

8. t'olrrl- i~oolulinrrrc~.\ i r i E" (rr 1 3) arc introduced hy thc equations

a) Show that if x = ( X I . . . . .r,,) and (.r,,-1. x,,) is not 10, 0 ) or of the form (c. 0 ) with c > 0, then unique polar coordinates are assigncd to x hy the requirements r = (rf ... ,Y:)'!~ > 0,O < < n . .. . . O < < r,0 < @,,-I < ~ T C (Hence . the limits of integration for a multiple integral over thc n-dimensional spherical region r'I . . + . r i 5 (1' arc 0 t o o for r. (3 to x for cach &, except 0 to 2 x for @,,. I .) [Hint: x cannot be 0. so .rf .. > 0 and r > 0 . Choosc @ I as cos-'(.uI / r ). O 5 @ I 5 r. = TC cannot occur. since they imply /.\I = r and hcnce Show that = 0 or .x! = .KT = . . . = 0. Next choose = cos-'(u2/(r sin41 )). 0 5 @2 5 I T , and show that 4L = 0 or 7r cannot occur. ctc.]

+ +

+.

+ . + .ri

b) Show that the corresponding Jaeobian is J,, = r"-' sin1'--' s i n n - '@! . . . sin 4 , ~ . [Hint: Use induction. For the step from n - 1 tori. use the polar coordinates p . @?. . . . , @,,-I in the subspacc xl = 0. so that x l = p cos 4:. .\ 7 = p sin @? cos @?. . . . . and

. 1 can be considered as cylindrical coordinates in El'. Go from thcsc cylindrical coordinates lo full polar coordinates by the equations .rl = r c o ~ @p ~=. r sin 41.Show that x 1 . p . 42% ...

whcrc J,,-l = p 11-1 s i n 1 ! 7 4,. now follows tiom (2.5 I).]

. . sin@,,1

by induction hypothesis. The conclusion

We recall the parametric representation for curves in the plane:

For each r the vector v = ( j ' ( r ) . g ' ( r ) ) can b e considered as the veloc,it,v \,ec.tor of the moving point ( r ( t ) , v ( t ) ) in terms of time t . Its magnitude is

Chapter 2 Differential Calculus of Functions of Several Variables where .r is distance along the path. One can interpret (2.85) as a vector function

where r is the position vector:

r = xi

+ yj.

Then

.

, cir dy v=i + -J = r ( t ) . dt clt The vector v is tangent to the path at the point (.r(t), y ( t ) ) . Normally, one can use s as a parameter, instead o f t , and then the velocity vector becomes

a litlit tangent vector, since by (2.86),

[For a detailed discussion, see CLA, pp. 329-336.1 The preceding analysis extends to curves in space with no significant change. A curve in space has parametric equations

and these are equivalent to a vector function r = r ( t ) , where now

as in Fig. 2.1 I . The velocity vector d v J. v =-id x + L dt dt

dz + -k

= rl(t)

dt

is again tangent to the path and has magnitude Ivl = d s l d t , where s is distance along the path, increasing with t. The corresponding unit tangent vector is 1

T =-v Ivl

Figure 2.1 1

dx ds

= -i

dz + -Jd v . + -k.

ds

ds

Curve in space, velocity vector v. and unit tangent vector T

123

124

Advanced Calculus. Fifth Edition

Figure 2.12

D ~ s p l a c e m e alo~lg ~ ~ t ~ the tangent Ilne

We remark that if v = 0 at a point. then the tangent direction is not determined by v; where this happens, it may be possiblc to determine a nonzero tangent vector by repeated differentiation (Problem 19). At a value t l o f t for which v is not 0, v = r l ( t l ) is a vector along the tangent line to the path at ( . t i . !.I. , - I ) = r ( t l ) , and hence the tangent line has parametric equations

or. in vector form,

(Section 1.3). We observe that Eq. (2.92) can be obtained from Eqs. (2.88) by taking differentials

and then interpreting dt as t - t l . du as .I- - . u l , dy as !, - yl, d: as ,- - ,-,.while evaluating the derivatives ,f"(t). . . . at t,. Thus d.u, r i y . d: are displacements from PI along the tangent line. as in Fig. 2.12. This result illustrates a basic principle. namely, that tnkitlg rl/fferet7tial.c. ~.,orrcspotlds t o JOrtning tLlngct~ts.Wc can write Eqs. (2.94) in vector form:

+

+

where d r = d,ui dyj drk. the "differential displacement vector." Now let an cquation

by given: in general this represents a surface in space. Let ( . v I , !I. this surface, and let

:,)

be a point on

Chapter 2

Differential Calculus of Functions of Several Variables

be a curve in the .sllrjuce, passing through (.rl. ) . I .:I

)

when t = r l . One has thus

Taking differentials. one concludes

where JF/J.r, ilF/ily, RF/Bz are to bc evaluated at (.rl, y , . :I) and d.1- = ,/"(tl) d t , dy = gl(tl)rlt, r l ~= h l ( t l ) d t : as earlier. these differentials can be replaced by .r - .rl y - y l . :- z I , where (.r. y. :) is a point on the tangent to the given curve at (.rI.y l . :I). Accordingly, (2.99) can be written:

.

This is the equation of a plane containing the tangent linc to the chosen curve. However, (2.100) no longer depends on the particular curve chosen: all tangent lines to curves in the surface through (.rl. y l . : I ) lie in the one plane (2.100). which is termed the rrznget~rplaneto the srr$ac.e a t (.rI,y l . :I). [If all three partial derivatives of F areoat the chosen point, Eq. (2.100) fails to determine a plane, and the definition breaks down.] It should be remarked that (2.99) is equivalent to (2.100). that is. that again the operation of rt~kit~g differentials yielrls the tatlgerzt. As a further illustration of this. consider a curve determined by two intersecting surfaces:

Thc corrcsponding differential relations

:,

represent two intersecting tangent planes at the poinl (-1-1.y l . ) at the point considered; thc intersection of these planes is thc trzngetlt lirzrj to the curve (2.101 ). To obtain thc equations in the usual form. the partial derivatives must be evaluated a1 (.rl.X I , ). and the differentials d.r, rly, d: must be replaced by 1---.\-I. y - yl. :-:I. These results can also be put in vector form. First of all. (2.100) shows that the vector (i)F/i).r)i (i)F/i)v)j (2FliI:)k is nor711nlto thc tangent plunc at ( . u l , j l , :I). This vector is known as the gratlietlt \lcjc.torof the function F(.r. y. :) (see Fig. 2.13). We write

+

+

i

grad F = -I

i1.r

.

i)F + -Ji;I)' . + -k. 0:

The notation V F (read "del F") for grad F will also be uscd; this is discussed in the following section. There is a gradient vector of F at each point of the domain of definition at which the partial derivatives exist: in particular, there is a gradient vector at each point of the surface F(.r, y , ,-) = 0 considered. Thus the gradient vector has a "point of application" and should be thought of as a holltici vector.

125

126

Advanced Calculus, Fifth Edition

Figure 2.13

Gradient vector of F and surface F ( r . y . z ) = const

Equation (2.99) for the tangent plane can now be written in the vector form grad F . d r = 0 ( r =.ri

+ y j + zk).

(2.104)

while the two Eqs. (2.102) for the tangent line become grad F . dr = 0.

grad G . d r = 0.

Since (2.105) expresses the fact that d r = dxi grad F and grad G, one concludes that

(2.105)

+ dyj + d z k is perpendicular to both

dr x (grad F x grad G ) = 0;

(2.106)

this equation again represents the tangent line. The vector grad F x grad G has components

Hence the tangent line can be written in the symmetric form X -XI

-

ac ,c

y

il G

- yl

ij G

-

Z -

ZI

;)F J

iJF

iJG

i3G

(2.108)

ity

or, in terms of Jacobians, in the form

This discussion and that of the preceding section show the significance of the differential. Taking differentials in an equation or system of equations corresponds on the one hand to replacement of the equations by linear equations in the variables d x , d y , . . . and on the other hand to replacement of curves and surfaces by tangent lines and planes.

Chapter 2

Differential Calculus of Functions of Several Variables

PROBLEMS 1. Given the equations x = sin t , J = cos 1, z = sin' r of a space curve. a) sketch the curvc, b) find the equations of the tangent line at the point P for which r = C)

$,

find the equation of a plane cutting the curvc at right angles at P .

2. a ) Show that the curve of Problem 1 lies in the surface .r' + 2>i2 + z = 2. b) Find the equation of the plane tangent to the surface at the point P of Problem l(b). c) Show that the tangent line to the curve at P lies in the tangent plane to the surface. 3. Let u = u(t ) and v = v(t ) be vector functions o f t , wherc u , v are vectors in 3-dimensional space. Prove (assuming appropriate differentiability):

d) z(1 ( ~X

V)

= U x dV dr

+ dr

x v

(watch the order!)

4. Let u = ~ ( t have ) constant magnitude, lu(t)l = a = const. Show. assuming appropriate differentiability. that u is perpendicular to d u l d t . What can be said of the locus of P such that

3 = u?

-+ 5. Let a point P move in space, so that OP = r = r(t) and the velocity vector is v = d r l d t . The accrlc~t-rrtior~ rtec.tor of P is dv dt

a=-=-

rl'r dt'

Assume further that the speed is 1 , so that t can be identified with arc length s and

a ) Show that v is perpendicular to the acceleration vector a = d v / d s (cf. Problem 4). The plane through P determined by v and a is known as the osculating plunr of the curve at P . It is given by the equation

wherc Q is an arbitrary point of the plane, provided that v x a # 0. Show that the osculating plane is also given by

+ dr

d'r PQ. -x-=O. dt dr2

where t is an arbitrary paramctcr along the curve, provided that ( d r l d r ) x (c12r/dt') # 0. b) Find the osculating plane of the curvc of Problen~1 at the point t = n / 3 and graph. [It can be shown that this plane is the limiting position of a plane through three points P I . P2. P3 on the curve, corresponding to parameter values tl. tz. t3, as the values r l , r7. t3 approach the value t at P. I

127

128

Advanced Calculus, Fifth Edition

6. Let I' move in space with spccd I as in Problcm 5. Thc c.lrr~~:trrr-c, of the path is defined as K . whc1.c

Thus if A- # 0. T and ( I / ~ ) r l T / r l s= N arc perpendicular unit vcctors: T is tangcnt to thc curvc at P . N is normal to the curvc at P and is termed thc prirrciptrl normal. The rrrrliu.~cfc-u~~trture is p = I / K . The vcctor B = T x N is called the hirrorn~ttl.Establish thc following relations:

a) B is a unit vcctor and (T. N. B) is a positive triple of unit vcctors

b) $ . ~ = O a n d g . ~ = O

2 = -rN;

C)

thcre is a scalar -s such that

d)

2 = - K T + r B. The equations

r

i4

known as thc mrsinr~

are known as the Frrneffortnlt1tl.v. For further properties of curves, see the book by Struik listed at the end of the chapter.

7. Let a point P move in space at speed v = ti.r/dt = ivl # 0. Show:

2

a) a = = $T+ $N ( p # 0,. [Hint: Sct v =vT, differentiate. and use the formula cfT/ds = K N = ( l / p ) N of Prohlcm 0.1 b ) K = L I - ~ ~ xV al. C)

If w = rla/dt and v x a # 0, then r = Iv x al--'v x a . w.

= 0.

d ) The path is a straight line if and only if

K

e) The path lies in a plane if and only iT s

= 0.

Note.

Wherc v = 0, the curvature

K

is defincd to bc 0. If v

= 0, thc path reduces to a

point.

8. For each of the following surfaccs, find the tangent plane and normal linc at thc point indicated. verifying that the point is in the surface: a ) .r' y' = 9 at (2.2, I ) b) r , : t b 2 - , - 2 -Oat(O,O, 1)

+ + ,-'

c ) .r3-.ry'+!:'-z3=0at(l. d) .u' + ?.' = 0 at (0,0. 0)

:'

1, 1 )

Why docs the procedure break down in (d)? Show by graphing that a solution is impossible.

e) .rJ -:= O a t ( . v ~ , y ~ . ; ~ ) . w h e r c . r=:I ~y~

+

f ) x y + v z + . \ z = I a t ( . r l . y ~ . z ~ ) . w h c r c . r ~ Vv J~: ~ +.r1;1

= 1

9. Show that thc tangcnt plane at (.ul. y ~: I.) to a surl'acc given by an cquation is as follows:

Obtain the cquatinns of thc normal linc.

z

= . f ( x . y)

Chapter 2

Differential Calculus of Functions of Several Variables

10. Find the tangent plane and normal line to the following surfaces at the points shown (see Problem 9):

+

a) z = x 2 y2 at (I, 1,2), b) z = at (:, c) z = at(2, 1,2),

Jw 3, f),

.

,

(i,

+

d) z = log(x2 y2) at $, 0). 11. For each of the following curves (represented by intersecting surfaces), find the equations of the tangent line at the point indicated, verifying that the point is on the curve: a) 2x y - z = 6 , x 2y 22 = 7 at (3, 1, 1): b) x2 y2 _f z2 = 9, x2 y2 - 8z2 = 0 at (2,2, 1); c) x 2 + y 2 = l , x + y + z = O a t ( l , O , - 1 ) ; d) x2 y2 z2 = 9, x 2 2y2 3z2 = 9 at (3,0,0). Why does the procedure break down in (d)? Show that solution is impossible. 12. Show that the curve

+ +

+ + +

+ +

+

+

x2-y

2

+ z 2 =1,

xy+xz=2

is tangent to the surface xyz - x 2 - 6 y = -6 at the polnt (1, 1, 1). 13. Show that the equation of the plane normal to the curve F ( x , y, z) = 0,

G(x, y , z) = 0

at the point ( x l , y l , zl) can be written in the vector form

3 "

dr . grad F x grad G = 0, and write out the equation in rcctangular coordinates. Use the results to find the normal planes to the curves of Problem 1I(b) and (c) at the points given. 14. Determine a plane normal to the curve x = t2, y = t , z = 2t and passing through the point (1.0, 0). 15. Find the gradient vectors of the following functions: a) F = x2

+ y2 + z2

b) F = 2 x 2 + y 2

Plot a level surface of each function and verify that the gradient vector is always normal to the level surface. 16. Three equations of form x = f (u, v), y = g(u, v), z = h(u, v) can be considered as parametric equations of a surface, for elimination of u and v leads in general to a single equation F ( x , y, z) = 0. a) Show that the vector

is normal to the surface at the point ( X I , yl, zl), wherexl = f ( u I , vl), yl = g(ur, VI), zl = h(u1, V I )and the derivatives are evaluated at this point.

129

130

Advanced Calculus, Fifth Edition b) Write thc cquation of the tangent plane. c) Apply thc results to find the tangent plane to the surfacc

at the point for which u = r / 4 , u = r / 4 .

d) Show that the surface (c) is a sphere. 17. Find the equations of a tangcnt line to a curve given by equations: z = f(x. y ) .

2 = s ( x . v).

18. Threc equations F(.x, y , Z , r ) = 0,

G(I, y . z , r ) = 0.

H(.Y,Y. 2 . 1 ) = 0

can bc regarded as implicit parametric equations of a curvc in terms of thc parameter r. Find the equations of the tangent line.

19. Let a curve in space be given in parametric form by an equation: r = r(r),tl 5 t 5 tz. whcre r = $ and 0 is fixed. a ) Show that if d r l d t = 0, then the curve degenerates into a single point. b) Let d r l d t = 0 at a point POon the curvc, at which t = to. Show that if d 2 r / d t 2 # 0 f o r t = to, then d 2 r / d t 2 represents a tangent vector to the curve at Po and, in general, that if

f o r t = to, then

represents a tangent vector to the curve at Po. [Hint: For t # to thc vector

represents a secant to the curve, as in Fig. 2.11. Now let t + ro and cvaluatc the limit of this vector with the aid of de l'H8pital's Rule, noting that f"(ro) = 0. gl(fO)= 0, hl(to) = 0, but at least one of the three (n 1)th derivatives is not 0.1

+

Show that if the parameter t is the arc length s, then d r l d s has magnitude 1 and hence is always a tangent vector. 20. Let a curve be given as in Problems 5-7. with v # 0 and u as parameter. Then thc cquation C)

givcs the tangent linc to the path at the point r(u) with w as parameter along the linc. Thc equation also defines a surfacc, with u and w as parameters. Thc surfacc is called thc tangential developable of the curve. It is a ruled surfacc whose rulings arc the tangent lines to the path. a ) Sketch the tangential developable for each of the following curves: (i) a circular helix; (ii) the curve x = u, y = u2, z = u'.

b) Show that along a chosen ruling of a tangential developable all points of the surface have the same tangent plane, which is the osculating planc to the curvc at the point r(u ).

Chapter 2 Differential Calculus of Functions of Several Variables Remark A developable sugace is one which can be obtained from a plane by bending without stretching. Examples are cones, cylinders, and the surfaces of this problem. (See Chapter 2 of the book by Struik listed at the end of the chapter.)

.:

f

Let F ( x , y , z ) be given in a domain D of space. To compute the partial derivative 13Fla.x at a point ( x , y , z ) of this domain, one considers the ratio of the change A F in the function F from ( x , y , z ) to ( x A x , y , z ) to the change A x in x . Thus only the values of F along a line parallel to the x axis are considered. Similarly, a F / a y and a F / a z involve a consideration of how F changes along parallels to the y and z axes, respectively. It appears unnatural to restrict attention to these three directions. Accordingly, one defines the directional derivative of F in a given direction as the limit of the ratio

+

of the change in F to the distance A s moved in the given direction, as A s approaches 0. Let the direction in question be given by a nonzero vector v. The directional derivative of F in direction v at the point ( x , y , z ) is then denoted by V,F ( x , y , z ) or, more concisely, by V, F . A displacement from ( x , y , z ) in direction v corresponds to changes A x , A y , A z proportional to the components v, , v , , v , ; that is,

Ax = hv,,

Ay = hv,,

Az = hv,,

(2.111)

where h is a positive scalar. The displacement is thus simply the vector h v , and its magnitude A s is h (vl.The directional derivative is now by definition the limit:

V,F

= lim h+O

F(x + h v , , y + h v , , z + h v , ) hlvl

F(x,y , z )

(2.112)

In this definition we have allowed h to approach 0 through positive or negative values, just as in the definition of the derivative d y l d x as the limit of A y l A x as A x approaches 0 through positive or negative values. One can think of the directional derivative of F at ( x , y , z ) in direction v as the derivative of a function of one variable obtained by considering F as a function of a coordinate along an axis through ( x , y , z) in the direction of v with the same unit of distance as along the x , y , and z axes. If F has a total differential at ( x , y , z ) , then, as in Section 2.6,

Thus

131

132

Advanced Calculus, Fifth Edition If h approaches zero, the last three terms approach 0, while the left-hand side approaches V,, F . One has thus the equation

Now v,/lvl, v,/lvl, vz/lvl are simply the components of a unit vector u in the direction of v and ire thus, by Section 1.2, the direction cosines of v: 1

u = -(v,i IvI

+ v,j + vzk) = cosai + c o s p j + cos yk.

Accordingly, (2.1 13) can be written in the following form:

One has thus the fundamental rule: The directional derivative of a function F(x, y , z ) is given by

where a , /3, y are the direction angles of the direction chosen. The right-hand side of (2.114) can be interpreted as the scalar product of the vector grad F = V F = (a F / a . ~ ) i (a F / a y ) j -t(a F/a=.)k and the unit vector u. Thus by Section 1.2 the directional derivative equals the component of grad F in the direction of v:

+

It is for this reason that the notation V,F is used for the directional derivative. In the special case when v = i, so that one is computing the directional derivative in the .X direction, one writes

similarly, the directional derivatives in the y and z directions are

There are other situations in which a partial derivative notation is used for the directional derivative. A common one is that in which one is computing the directional derivative at a point (x, y , z ) of a surface S along a given direction normal to S. If n is a unit normal vector in the chosen direction, one writes

Chapter 2 Differential Calculus of Functions of Several Variables Equation (2.115) gives further information about the vector V F = grad F . For from (2.1 15), one concludes that the directional derivative at a given point is a maximum when it is in the direction of V F ; this maximum value is simply

Thus the gradient vector points in the direction in which F increases most rapidly, and its length is the rate of increase in that direction. If v makes an angle 8 with VF, then the directional derivative in the direction of v is Thus if v is tangent to a level surface at (x, y, z), then V ,F = 0; for V F is normal to such a level surface. One sometimes refers to a "directional derivative along a curve." By this is" meant the directional derivative along a direction tangent to the curve. Let the curve be given in terms of arc length as parameter:

and let the direction chosen be that of increasing s. The vector

is then tangent to the curve and has length 1 (cf. Section 2.13). Hence the directional derivative along the curve is

that is, it is the rate of change of F with respect to arc length on the curve. The meaning of the directional derivative can be clarified by visualizing the following experiment. A traveler in a balloon carries a thermometer. At intervals he records the temperature. If his reading at position A is 42" and that at the position B is 44", he would estimate that the directional derivative of the temperature in the direction %is posi%e and has a value given approximately by 2" t d, where d is the distance IAB 1; if 1 AB 1 = 5000 ft, the estimate would be 0.0004" per foot. If he continued traveling past B in the same direction, he would expect the temperature to rise at about the same rate. If the balloon moves on a curved path at a known constant speed, the traveler can compute the directional derivative along the path without looking out of the balloon. The foregoing discussion has been given for three dimensions. It can be specialized to two dimensions without difficulty. Thus one considers a function F(x, y) and its rate of change along a given direction v in the xy-plane. If v makes an angle cr with the positive x axis, one can write

133

134

Advanced Calculus, Fifth Edition

Figure 2.14

Gradient of F ( x , y ) and curves F ( x , y ) = const.

v

for v has direction cosines cos a and cos /? = cos (;n - a ) = sin a . Again the directional derivative is the component of grad F = - ( a ~ / a x ) i ( a F / a y ) j in the given direction; the directional derivative at a given point has its maximum in the direction of grad F , the value being

+

The directional derivative is zero along a level curve of F , as suggested in the accompanying Fig. 2.14. If the level curves are interpreted as contour lines of a landscape, that is, of the surface z = F ( x , y ) , then the directional derivative means simply the rate of climb in the given direction. The rate of climb in the direction of steepest ascent is the gradient, precisely the term in common use. The bicyclist zigzagging up a hill is taking advantage of the component rule to reduce the directional derivative.

PROBLEMS 1. Evaluate the directional derivatives of the following functions for the points and directions given: a) F ( x , y , z ) = 2 x 2 - y2 z2 at (1, 2 , 3 ) in the direction of the line from (1, 2, 3 ) to (3, 5,O); b) F ( x , y , z ) = x2 y2 at (O,O, 0 ) in the direction of the vector u = ai bj +ck; discuss the significance of the result; c) F ( x , y ) = ex cosy at ( 0 . 0 ) in a direction making an angle of 60" with the x axis; d) F ( x , y ) = 2x - 3 y at ( 1 , 1 ) along the curve y = x 2 in the direction of increasing x; e) F ( x , y , z ) = 3x - 5 y 22 at ( 2 , 2 , 1 ) in the direction of the outer normal of the surface x2 y2 z2 = 9; f ) F ( x , y , z ) = x2+y2-z2at(3,4,5)alongthecurvex2+y2-z2 = 0 , 2 x 2 + 2 y 2 - z 2 = 25 in the direction of increasing x ; explain the answer.

+

+

+

+ +

+

Chapter 2 Differential ~ a l c u l u of s Functiuns d Several Variables

2. Evaluate a F / a n , where n is the unit outer normal to the surface given, at a general point (x, y , z ) of the surface given: a) F = x 2 - y2, surface x 2 + y2 + z2 = 4 ; b) F = x y z , surface x 2 + 2 y 2 4z2 = 8. 3. Prove that if u = f (x, y ) and v = g(x, y ) are functions such that au av au - a v - - -ax ay' ay ax' then . 1cVcx~= V a + n / ~ v *. .r, for every angle a. 4. Show that if u = f (x, y ) , then the directional derivatives of u along the line 0 = c d w and circles r = const (polar coordinates) are given respectively by

+

a

~

,-.

-\

- 9

5. Show that under the hypotheses of Problem 3 one has au ar

--

6.

7.

8. 9.

1 av r ae'

--

I r

a~ ae

--

a~ - --

ar' [Hint: Use Problem 4.1 Let arc length s be measured from the point (2,O) of the circle x 2 y2 = 4 , starting in the direction of increasing y . If u = x2 - y2, evaluate d u l d s on this circle. Check the result by using both the directional derivative and the explicit expression for u in terms of s. At what point of the circle does u have its smallest value? Under the hypotheses of Problem 3 , show that & / a s = a v / a n along each curve C of the domain in which u and v are given, for appropriate direction of the normal n. Determine the points ( x , y ) and directions for which the directional derivative of u = 3x2 y2 has its largest value, if (x, y) is restricted to lie on the circle x2 y2 = 1. A function F(x, y , 2)is known to have the following values : F(1, 1, 1) = 1, F(2, 1, 1) = 4 , F(2,2, 1 ) = 8, F(2, 2 , 2 ) = 16. Compute approximately the directional derivatives:

+

+

+

V i F , V , + j F , Vi+j+kF at the point ( I , 1, 1).

Let a function z = F ( x , y ) be given; its two partial derivatives a z / a x and a z / a y are themselves functions of x and y:

Hence each can be differentiated with respect to x and y; one thus obtains the four second partial derivatives:

135

136

Advanced Calculus, Fifth Edition Thus a2z/ax2is the result of differentiating a z / a x with respect tox, while a2z/ay ax is the result of differentiation of a z / a x with respect to y. Here a simplification is possible if all derivatives concerned are continuous in the domain considered, for one can prove that

that is, the order of differentiation is immaterial. A proof of (2.124) is given at the end of this section. Third- and higher-order partial derivatives are defined in a similar fashion, and again, under appropriate assumptions of continuity, the order of differentiation does not matter. Thus one obtains four third partial derivatives:

A clue as to why the order of differentiation does not matter can be obtained by considering the mixed partial derivatives of z = x n y m .Here one has

Thus the order does not matter in this case. A similar reasoning applies to a sum of such terms with constant factors, that is, a polynomial in x and y:

It is essentially because an "arbitrary" function can be approximated by such a polynomial near a given point that the rule holds. Other notations for higher derivatives are illustrated by the following examples:

a2z = z x x = f l l ( ~Y, ) , ax2

a2z

-= Z y x = f21(x, Y ) , ax a y

If z = f ( x , y), the Laplacian of z , denoted by Az or v 2 z , is the expression

The A symbol here must not be confused with the symboI for increments; for this reason the v2znotation is preferable. If w = f (x, y, z), the Laplacian of w is

Chapter 2 Differential Calculus of Functions of Several Variables defined analogously:

The origin of the v2symbol lies in the interpretation of V as a "vector differential operator":

One then has symbolically

I

This point of view will be discussed further in Chapter 3. If z = f (x, y) has continuous second derivatives in a domain D and

v2z= 0

(2.126)

in D, then z is said to be harmonic in D. The same term is used for a function of three variables that has continuous second derivatives in a domain D in space and whose Laplacian is 0 in D. The two equations for harmonic functions:

are known as the Laplace equations in two and three dimensions, respectively. Another important combination of derivatives occurs in the biharmonic equation:

which arises in the theory of elasticity. The combination that appears here can be expressed in terms of the Laplacian, for one has

If we write v 4 z = v 2 ( v 2 z ) ,then the biharmonic equation can be written:

Its solutions are termed bihamzonic functions. This can again be generalized to functions of three variables, (2.129) suggesting the definition to be used. Harmonic functions arise in the theory of electromagnetic fields, in fluid dynamics, in the theory of heat conduction, and in many other parts of physics; applications will be discussed in Chapters 5, 8, and 10. Biharmonic functions are used mainly in elasticity. (See Section 8.25.)

137

138

Advanced Calculus, Fifth Edition Proof of the rule (2.124). We assume that z = f ( x , y) has continuous derivatives fx, fy, fxy , f v x in a domain D. Let R be a closed square region xo 5 x 5 xo + h, yo 5 y Iyo h contained in D. Then by the rule gf(x)dx = g ( b ) - g(a),

+

// tr

L~

dx dy =

l;+h ll:+h fx."

dy dx

-

,,

I

,.,.,

>

R

= f (xo

+ h , YO + h ) - f'(xo, YO + h ) - f (xo + h , yo) + f (xo, yo).

We interchange the roles of x and y to obtain similarly

Therefore the two double integrals are equal or

Since this holds for every such square region R in D, we must have fx,

(x, Y) = f,, (x, y)

in D.

(See Problem 14 following Section 2.18.) Thus the proof is compete. Remark. This proof uses properties of double integrals which are usually developed in the first course in calculus; these properties are also considered in Chapter 4 (Sections 4.3 and 4.11).

Let z = f (x, y) and x = g(t), y = h(t), so that z can be expressed in terms of r alone. The derivative d z l d t can then be evaluated by the chain rule (2.33) of Section 2.8:

By applying the product rule. one obtains the following expression for the second derivative:

139

Chapter 2 Differential Calculus of Functions of Several Variables

To evaluate the expressions ( d / d t ) ( a z / a x )and ( d / d t ) ( a z / a y ) ,one uses (2.130) again, this time applied to a z l a x and a z / a y rather than to z.

One thus finds the rule:

This is a new chain rule. Similarly, if z = f ( x , y ) , x = g ( u , v ) , y = h ( u , v ) , so that (2.34) holds, one has

az -- -a z ax -

au

ax

au

az ay + --, ay au

a2z - az a2x - - - - + - - -a+( -a z )+ax- - - az a2y au2

ax au2

au

ax

au

ayau2

(2.132)

a ( a z ) ay au ay a ~ ' i

*

Applying (2.132) again, one finds

so that

It should be remarked that (2.133) is a special case of (2.13 l), since v is treated as a constant throughout. Rules for a 2 z / a u a v , a 2 z / a v 2 and for higher derivatives can be formed, analogous to (2.13 1) and (2.133). These rules are of importance, but in most practical cases it is better to use only the chain rules (2.33), (2.34), (2.35), applying repeatedly if necessary. One reason for this is that simplifications are obtained if the derivatives occurring are expressed in terms of the right variables, and a complete description of all possible cases would be too involved to be useful. The variations possible can be illustrated by the following example, which concerns only functions of one variable. Let y = f ( x ) and x = e'. Then

140

Advanced Calculus, Fifth Edition Hence

"[

One could also write

:

1

, dy d x e = X -dx

dy-- dy dt

and then

The second method is clearly simpler than the first; the answers obtained are equivalent because of the equation x = et.

2.17 THELAPLACIAN IN POLAR, CYLINDRICAL, AND SPHERICAL COORDINATES An important application of the method of the preceding section is the transformation of the Laplacian to its expression for other coordinate systems. We consider first the 2-dimensional Laplacian

and its expression in terms of polar coordinates r, 0. Thus we are given w = f ( x , y) and x = r cos6, y = r sine, and we wish to express V ' W in terms of r, 8, and derivatives of w with respect to r and 8. The solution is as follows. One has

by the chain rule. To evaluate arlax, aelax, arlay, aelay we use the equations dx =cosOdr -rsinOdO,

dy = s i n 8 d r + r c o s O d O .

These can be solved for d r and d e by determinants or by elimination to give dr=cos8dx+sinOdy,

sin 0 d6=--dx+-dy. r

cos (3 r

a0 ax

---

Hence ar ax

- = cos8,

ar ay

- = sine,

--

sin 8 --

r '

a0 ay

C O S ~

r '

i

Chapter 2 Differential Calculus of Functions of Several Variables Thus (2.134) can be written as follows:

These equations provide general rules for expressing derivatives with respect to x or y in terms of derivatives with respect to r and 8 . By applying the first equation to the function a w l a x , one finds that

by (2.135) this can be written as follows:

P

The ruIe for differentiation of a product gives finally a2w 2 s i n 6 c o s 8 a 2 w a2w = COS 0-

ax~

ar2

r

ar a6

+--sin20a2w r 2 ae2

In the same manner, one finds a2w

a

aw

p = , ( , )

=sine-:r(

aw + C-O S aw ~ r a ,

sine-

Adding (2.136) and (2.137), we conclude:

this is the desired result. Equation (2.1 38) at once permits one to write the expression for the 3-dimensional Laplacian in cylindrical coordinates; for the transformation of coordinates

involves only x and y and in the same way as earlier. One finds:

142

Advanced Calculus, Fifth Edition

9

A procedure similar to the preceding gives the 3-dimensional Laplacian in spherical coordinates (Fig. 2.10): v2", = lrl l

a2w a2w a2w aZw + -ia2w I a2w ---++ ---=+ ax2 ay2 a22 ap2 p2 a@ ~2 sin2# a02 2aw +--+-P aP

(2.140)

C O ~ # ~ W

P2

ad'

(See Problem 8 following the next section.)

In Section 2.10, procedures are given for obtaining differentials or first partial derivatives of functions implicitly defined by simultaneous equations. Since the results are in the form of explicit expressions for the first partial derivatives [cf. (2.71), (2.72), (2.73), etc.], these derivatives can be differentiated explicitly. An example will illustrate the situation. Let x and y be defined as functions of u and v by the implicit equations:

a ( F , G)

ax - - - = m a, - a ( F , G )

TX-3

Hence

PROBLEMS 1. Find the indicated partial derivatives:

~ w. = ---I and aay-2 ~f JW'

a)

a2u1

b)

2 and u

dx-

i f w =arctan

f,

:-I ::/ 2x

2y

- -3, X

Chapter 2 c)

3 & and

Differential Calculus of Functions of Several Variables 1

.)

if w = eX--y-,

~ I I I + # I ~ , ,

1f w = xmyn.

d)

Formula (2.138) can be used as a check for (a) and (b). 2. Verify that the mixed derivatives are identical for the following cases: a)

w. and & forz = *, + a22

b) a

,Z

x-

n , m. and

a7w

a3w

aJu,

y

for w = Jx2

+ y2 + z2.

3. Show that the following functions are harmonic in x and y: a) ex cos y, b) x3 - 3xy2,

C ) log 4. a) Show that every harmonic function is biharmonic. b) Show that the following functions are biharmonic in x and y:

xeXcosy,

Jm.

x4 - 3x2y2.

c) Choose the constants a , b, c so that ax2

+ bxy + cy2 is harmonic. + bx2y + cxy2 + dy3 is harmonic.

d) Choose the constants a , b, c, d so that ax3

5. a) Prove the identity:

v2(uv)= u v 2 v

+ vv2u +2 v u . v v

for functions u and v of x and y. b) Prove the identity of (a) for functions of x , y, and z. c) Prove that if u and v are harmonic in two or three dimensions, then

is biharmonic. [Hint: Use the identity of (a) and (b).] d) Prove that if u and v are harmonic in two or three dimensions, then

+ +

6. 7. 8.

9.

is biharmonic, where r 2 = x2 + y2 for two dimensions and r 2 = x2 y2 z2 for three dimensions. Establish a chain rule analogous to (2.133) for a2z/au av. Use the rule (2.133), applied to a2w/ax2 and a2w/ay2,to prove (2.138). Prove (2.140). [Hint: Use (2.139) to express v 2 w in cylindrical coordinates; then note that the equations of transformation from ( z , r ) to ( p , 4) are the same as those from (x, y) to (r, 01.1 Prove that the biharmonic equation in x and y becomes

in polar coordinates (r, 8). [Hint: Use (2.138).] 10. If u and v are functions of x and y defined by the equations xy+uv=l, find a2u/ax2.

xu+yv=l,

144

Advanced Calculus, Fifth Edition

11. If u and v are inverse functions of the system show that u is harmonic. 12. A d~fferentialequation is an equation relating one or more variables and their derivatives; the Laplace equation and biharmonic equation above are illustrations (see Chapters 9 and 10). A basic tool in the solution of differential equations, that is, in determining the functions that satisfy the equations, is that of introducing new variables by appropriate substitution formulas. The introduction of polar coordinates in the Laplace equation in Section 2.17 illustrates this. The substitutions can involve independent or dependent variables or both; in each case, one must indicate which of the new variables are to be treated as independent and which as dependent. Make the indicated substitutions in the following differential equations: 'a)

2 = 4; new variables y (dep.) and u = x 2 (indep.); Y +x

dy

b)

z = 2r.+,Y-4

C)

x 2d" ;fr;i

+

' new variables v = y - 3 (dep.) and u = x - 1 (indep.); ;

d?. d + 3 x d& + 2 = 0; new variables y (dep.) and t = log x (indep.);

+

y 3)= 0; new variables x (dep.) and y (indep.); d) ;d2y i ; ~ (d E a)

d2

f ) a auz

- 44

+

2 + y(3x2 - 2) = 0; new variables u = e-''v

(dep.) and x (indep.);

BU

bq = 0 ( a , b constants); new variables u (dep.), z = bx - a y (indep.), and w = a x by (indep.);

+

,x,- p = 0; new variables u (dep.), z = x + y (indep.), and w = x - y (indep.). 13. Transformation of the Sturm-Liouville differential equation. The equation is g)

on an interval a 5 x 5 b on which p(x) > 0, k(x) > 0 and h is a constant. One transforms this by the substitution u = z(x)v, y = h(x), where h'(x) # 0 and z(x) # 0 on the interval, to obtain a new dcpendent variable v and new independent variable y. a) Show that the new equation is d2v dv p ~ h ' ~ - (pzh" 2pz'h' plzh)-- f [pz" p'z' z(hk - q)]v = 0, dy2 d.~ where in all coefficients x is expressed in terms of y through the inverse function x = 4(y) of y = h(x). (Assume all functions and derivatives appearing are continuous.) b) Show that for proper choice of h(x) and z(x) the new equation can be written thus:

+

+

+

+

+

[Hint: To obtain the desired form, require that the coefficient of d v l d y is 0 and that the coefficient of hv equals the coefficient of d2v/dy'; then divide by this latter coefficient. Show that the second condition gives h' = (k/p)'/*, so that a suitable function h is g,[k(t)/p(t)]1/2dt, where xo is on the given interval, and that the first condition gives ( d l d x ) log(h'z2p) = 0, which is satisfied by a suitable function by making h'z2p = 1 and then choosing z = (kp)-'I4. Verify that the Q(v) obtained can be written as (l/f)d"/dy2 q l k , where f = i / z as a function of y. This change of variables is known as the Liouville substitution.]

+

Chapter 2 Differential Calculus of Functions of Several Variables 14. The proof at the end of Section 2.15 uses properties of the double integral. Justify the following: a) If g(x, y) is continuous on a square region R: xo 5 x 5 xo h , yo 5 y 5 yo h, and g(x,y ) > 0 in R, then

+

+

b) If g ( x , y) is continuous in domain D and

for every region R as in (a) contained in D, then g(x,y) GO. [Hint: Suppose g(xo,yo) > 0. Then by continuity, g(x,y) > 0 in some region R (Problem 7 following Section 2.4). Hence by (a) the double integral over R could not be 0. If g(xo,yo) < 0, then apply the same reasoning to -g(x, y).]

AND MINIMAOF FUNCTIONS 2.19 MAXIMA OF SEVERAL VARIABLES

We first recall the basic facts concerning maxima and minima of functions of one variable. Let y = f (x) be defined and differentiable in a closed interval a _( x _( b, and let x o be a number between a and b: a < xo < b. The function f (x) is said to have a relative maximum at xo i f f (x) 5 f (xo)for x sufficiently close to xo.It follows from the very definition of the derivative that i f f '(xo)> 0, then f ( x ) > f (xo)for all x > xo and sufficiently close to xo; similarly, if f '(xo) < 0, then f (x) > f (xo)for all x < xo and sufficiently close to xo. Hence at a relative maximum, necessarily f '(xo)= 0. A relative minimum of f (x) is defined by the condition f (x) f (xo) for all x sufficiently close to xo. A reasoning such as the preceding enables one to conclude that at a relative minimum o f f (x),necessarily f '(xo) = 0. . The points xo at which fl(xo) = 0 are termed the critical points of f ( x ) . Although every relative maximum and minimum occurs at a critical point, a critical point need not give either maximum or minimum. This is illustrated by the function y = x 3 ,- 1 ( x _( 1, at x = 0. This function has a critical point at x = 0, but the point is neither maximum nor minimum and is an example of what is termed a horizontal injection point. This is illustrated in Fig. 2.15. Let xo be a critical point, so that f '(xo)= 0,and let it be assumed that f "(xo)> 0. Then f (x) has a relative minimum at xo. For by the Mean Value theorem when x > xo, f (x)- f (xo)= f '(XI)(x - XO), where xo t xl < x. Since f "(xo) > 0, fl(x,)> 0 for X I =. xo and X I sufficiently close to xo. Hence f (x) - f (xo)> 0 for x > xo and x sufficiently close to xo. Similarly, f (x) - f (xo) > 0 for x < xo and x sufficiently close to xo. Therefore f (x)has a relative minimum at xo. An analogous reasoning applies when f "(xo) < 0. Accordingly, one can state the rules: ,

If f '(xo) = 0 and f "(xo) > 0, then f (x)has a relative minimum at xo; i f f '(xo) = 0 and f "(xo) < 0, then f (x) has a relative maximum at xo.

145

146

Advanced Calculus, Fifth Edition Horizontal ~nflect~on polnt mln

Figure 2.15

Critical points of function y = f ( x ) .

These two rules cover most cases of interest. It can of course happen that f "(xo) = 0 at a critical point, in which case one must consider the higher derivatives. A repeated application of the previous reasoning gives the following general rule: Let f ' ( x O ) = 0, f ll(xO)= 0 , . . . , f ( n ) ( ~ o=) 0, but f ("+')(xo) # 0; then f ( x ) has a relative maximum at xo if n is odd and f '"+"(xo) < 0; f ( x ) has a relative minimum at xo if n is odd and f '"+"(xo) > 0; f ( x ) has neither relative maximum nor relative minimum at xo but a horizontal inflection point at xo if n is even. The preceding discussion has been restricted to points xo within the interval of definition of f ( x ) and to relative maxima and minima. The notion of relative maximum and minimum can readily be extended to the endpoints a and b, and rules can be formulated in terms of derivatives. However, the interest in these points arises mainly in connection with the absolute maximum and minimum of f (x) on the interval a 5 x 5 b. A function f ( x ) is said to have an absolute maximum M for a certain set of values of x if f ( x o ) = M for some xo of the set given and f ( x ) 5 M for all x of the set; an absolute minimum is defined similarly, with the condition f ( x ) 1 M replacing the condition f ( x ) 5 M . The following theorem is then fundamental. THEOREM If f ( x ) is continuous in the closed interval a 5 x Ib, then f ( x ) has an absolute minimum MI and an absolute maximum M2 on this interval.

The proof of this theorem requires a more profound analysis of the real number system; see Section 2.23. It should be remarked that the inclusion of the end-values is essential for this theorem, as the simple example y = x for 0 < x < 1 illustrates; here the function has no minimum or maximum on the set given. For the set 0 5 x 5 1, the absolute minimum is 0, for x = 0, and the absolute maximum is 1, for x = 1. The example y = tanx for -3712 < x < n / 2 illustrates a function with no absolute minimum or maximum; in this case, adjoining values at the endpoints x = h / 2 does not help. A function f ( x ) is said to be bounded for a given set of x if there is a constant K such that I f ( x ) l 5 K on this set. The theorem above implies that i f f ( x ) is continuous in the closed interval a 5 x 5 b, then f ( x ) is bounded in this interval; for K can 1, J M 2 JFor . example, let y = sin x for 0 5 x 5 n. be chosen as the larger of JM1

Chapter 2 Differential Calculus of Functions of Several Variables This function has the absolute minimum M I= 0 and the absolute maximum M2 = 1 and is bounded, with K = 1. The example y = x for 0 < x < 1 illustrates a bounded function (K = 1) having neither absolute minimum nor absolute maximum; the example y = tanx for -in < x < in illustrates an unbounded function. To find the absolute maximum M of a function y = f (x), differentiable for a 5 x 5 b, one can reason as follows: if f (xo) = M and a < xo < b, then f ( x ) has necessarily a relative maximum at xo; thus the absolute maximum occurs either at a critical point within the interval or at x = a or x = b. One therefore first locates all critical points within the interval and compares the values of y at these points with the values at x = a and x = b ; the largest value of y is the maximum sought. It is thus not necessary to use the second derivative test described at the beginning of this section. The absolute minimum can be found in the same way. The determination of critical points and their classification into maxima, minima, or neither are important also for graphing functions; from a knowledge of the critical points and the corresponding values of y , a very good first approximation to the graph of y = f ( x ) can be obtained. After these preliminaries we can consider the analogous questions forfunctions of two or more variables. Let z = f ( x , y ) be defined and continuous in a domain D. This function is said to have a relative maximum at ( x o ,yo) if f ( x , y ) If (xo, yo) for ( x , y ) sufficiently close to (xo, yo) and to have a relative minimum at (xO,yo) if f ( x , y ) >_ f (xo, yo) for ( x , y ) sufficiently close to (xo, yo). Let (xo, yo) give a relative maximum of f ( x , y ) ; then the function f ( x , yo), which depends on x alone, has a relative maximum at xo, as is illustrated in Fig. 2.16. Hence if fx(xo, yo) exists, then fx(xo, yo) = 0; similarly, if f,(xo, yo) exists, then f y ( x o , yo) = 0. Points ( x , y ) at which both partial derivatives are 0 are termed critical points o f f . As before, one concludes that every relative maximum and every relative minimum occur at a critical point off, if f, and f, exist in D. One might expect that the nature of a critical point could be determined by studying the functions f ( x , yo) and f (xa, y ) , with the aid of the second derivatives as previously. First of all, it should be remarked that one of these functions can have a maximum at (xO,yo), while the other has a minimum. This is illustrated by the s has a minimum with respect to x at function z = 1 x 2 - y2 at (0,O); t h ~ function x = 0 for y = 0 and a maximum with respect to y at y = 0 for x = 0, as shown in

+

I I

I I

I /'YO I /

- - - - -Y - - - - - A

X

Figure 2.16

Maximum of z = f ( x , y).

1-Y

147

148

Advanced Calculus, Fifth Edition

Figure 2.17

Saddle point (z = 1 + x 2 - y 2 ) .

Figure 2.18

z = 1 - x 2 + 4 x y - y2.

Fig. 2.17. This critical point is an example of what is termed a saddle point; this will be discussed further. The level curves of z are shown in Fig. 2.17; the configuration shown is typical of that at a saddle point. A further complication is the following: It can happen that z = f (x,yo) has a relative maximum for x = xo, and z = f (xo,y) has also a relative maximum for y = yo, but z = f (x,y) does not have a relative maximum at (x0,yo). This is illustrated by the function z = 1 - x 2 4xy - y2 at (0,O).For y = 0, z = 1 - x2, with maximum for x = 0; for x = 0, z = 1 - y2, with maximum for y = 0.On the other hand, for y = x , z = 1 2x2, so that the section of the surface by the plane y = x has a minimum at x = 0. This is shown in Fig. 2.18. The level curves, shown in the same figure, again reveal the presence of a saddle point.

+

+

Chapter 2 Differential Calculus of Functions of Several Variables

Figure 2.19

a 2 z / a r 2for functlon of Fig. 2.18.

rl $3

A better understanding of this example and a clue to a general procedure can be obtained by introducing cylindrical coordinates (r, 0). One has then z = 1 r2(1- 2 sin 20). If one sets 0 = const = a, one obtains z as a function of r . For a = 0, one obtains thexz-trace: z = 1-r2; for a = n/2, one obtains the yz-trace: z = 1-r2; for a = x/4, one obtains the trace in the plane y = x : z = 1+r2. For a general a , one has the trace z = 1 - r2(1 - 2 sin 2a); r can here be allowed both positive and negative values. On each trace, az/ar = -2r(l- 2 sin 2a); thus z, as a function of r , has a critical point for r = 0. However, a2z/ar2 = -2(1- 2 sin 2a). This second derivative takes on both positive and negative values, as shown graphically in Fig. 2.19. The second derivative test for functions of one variable at once shows that both relative maxima and relative minima occur in corresponding directions a . In order to generalize this analysis to an arbitrary function z = f (x, y) we make use of the directional derivative in direction a , defined in Section 2.14. This we again denote by V,z and recall the formula:

At a critical point (xo, yo), one has necessarily V,z = 0; this means that in each plane ( x - xo) sin a - (y - yo) cos a = 0, z has a critical point at (xo, yo), when z is regarded as a function of "directed distance" s from (xo, yo), in the xy-plane; this is illustrated in Fig. 2.20. As in the example above, the type of critical point may vary with the direction chosen. In order to analyze the type, one introduces the second derivative of z with respect to s , that is, the second directional derivative in direction a. This is simply the quantity V,V,z, and one has

150

Advanced Calculus, Fifth Edition

Figure 2.20

Critical point of z = f ( x , y) analyzed by vertical sections.

This is precisely the quantity a2z/ar2of the preceding example. In order to guarantee a relative maximum for z at (xo, yo), one can require that this second derivative be negative for all a, 0 5 a 5 2n; this ensures that z has a relative maximum as a function of s in each plane in direction a, and in fact (all second derivatives being assumed continuous) that z has a relative maximum at the point; cf. Problem 10 following Section 2.2 1. A similar reasoning applies to relative minima. One therefore obtains the rule: If a z l a x = 0 and azlay = 0 at (.xO,yo) and

for x = xo, y = yo, and all a: 0 5 a 5 2n, then z = f (x, y) has a relative maximum at (xo, yo); if azlax = 0 and az/ay = 0 and a2z

-c0s2, ax2

+ 2-ax ay i n a cos u + 7 a2z sin a s 0 ay-

forx = xo, y = yo, and all a: 0 5 a 5 27r, then z = f (x, y) has a relative minimum at (xo, YO). It thus appears that the study of the critical points is reduced to the analysis of the expression

where the abbreviations

a2z

A = l(xo,

ax

yo),

a2z B = -(xo, YO), ax ay

a2z

C = --(xo,

ay2

yo)

(2.145)

are used. Here an algebraic analysis reduces the question to a simpler one, for one has the following theorem:

+

THEOREM If B~ - AC t 0 and A C < 0, then the expression (2.144) is negative for all a; if g2- AC < 0 and A + C > 0, then the expression (2.144) is positive for all a.

I

Chapter 2 Differential Calculus of Functions of Several Variables Proof. Let the expression (2.144) be denoted by P(a). Let B~ - AC < 0 and A C < 0. Then P(*lr/2) = C < 0, for if C 1 0, then A C < 0 implies A < 0, so that AC 5 0; this contradicts the condition B2 - AC < 0. Similarly, one shows P(0) = A <: 0. For a # &n/2, one has

+

+

P ( a ) = cos2a(A

+ 2B tan a + C tan2 a).

Thus P ( a ) is positive, negative, or 0, according to whether the quadratic expression Q(u) = c u 2

+ 2Bu + A

(u = tana)

is positive, negative, or 0. Since B~ - AC < 0, Q(u) has no real roots; thus Q(u) is always positive or always negative. For u = 0, Q = A < 0. Hence Q(u) is always negative, and the same holds for P(a). Accordingly, the first assertion is proved. The second is proved in the same way. If B2 - AC > 0, the proof just given shows that P ( a ) will be positive for some values of a and negative for others, as in the example of Fig. 2.17. In this case the critical point is termed a saddle point. If B 2 - AC = 0, there will be some directions in which P ( a ) = 0, and one must introduce higher derivatives in order to determine the nature of the critical point. We summarize the results obtained: E S, THEOREM Let z = f ( x , y) be defined and have continuous first and second partial derivatives in a domain D. Let (x0, yo) be a point of D for which azlax and az/ay are 0. Let

Then one has the following cases: B~ - AC < 0 and A + C < 0, relative maximum at (xo, yo);

(2.146)

B2 - AC < 0 and A + C > 0, relative minimum at (xo,yo);

(2.147)

B2 - AC > 0, saddle point at (xo, yo);

(2.148)

B2 - AC = 0, nature of critical point undetermined.

(2.149)

If A , B, and C are all 0, so that P ( a ) = 0, one can study the critical point with the aid of the third derivative: V a V a V a z .This is illustrated by the function z = x3 - 3xy2, whose level curves are graphed in Fig. 2.21. If the third derivatives are all 0, one can go on to higher derivatives as for functions of one variable. In essence the whole problem is reduced to one for functions of one variable, and the directional derivative is the chief tool in this reduction. In the accompanying Fig. 2.21, examples are given of various types of critical points; in each case the level curves of the function are shown. Remark. For the cases B~ - AC < 0 and B2 - AC > 0, one can show that the level curves o f f have the qualitative appearance of Fig. 2.21(a), (b), and (c). To this end,

151

I'

152

Advanced Calculus, Fifth Edition

(a) Maximum Y

(d) Multiple saddle point

11.

(b) Minimum

, i.

(c) Saddle point

Y

(e) Minimum

(f) Maximum

Figure 2.21 Examples of critical points of f (x, y ) at (0,O).The corresponding functions are as follows: (a) z = 1 - - y2, (b) = + y2, (c) z = x y , (d) z = X' - 3 x y 2 , (e) z = x Z y 2 , (f) z = 4 - y 2 .

one applies the implicit function theorem to the equation

f(x,y)-c=O to obtain a differential equation for these level curves:

The differential equation has a singularpoint at the critical point, since the numerator and denominator are 0 there. By applying the theory of configurations at such singular points, one deduces that in the case of a maximum or minimum, the level curves are all closed curves around the singular point, whereas at a saddle point the curves resemble hyperbolas and asymptotes as in Fig. 2.21(c). For details, see Chapter 15 of the book by Coddington and Levinson listed at the end of the chapter. This book considers the problem for pairs of first-order equations. Our problem is replaced by such a pair by writing

which gives the solutions in parametric form. The preceding discussion has concerned only relative maxima and minima. Just as with functions of one variable, one can define the notions of absolute maximum and absolute minimum; again their investigation calls for a study of the function at the "ends" of the domain D of definition, that is, on the boundary. In order to

Chapter 2 Differential Calculus of Functions of Several Variables ensure the existence of absolute maximum and minimum, one must demand that the domain D itself be bounded (cf. Section 2.2). One has then, as for functions of one variable: THEOREM Let D be a bounded domain of the xy-plane. Let f (x , y) be defined and continuous in the closed region E formed of D plus its boundary. Then f (x, y) has an absolute maximum and an absolute minimum in E.

The fact that D is a domain is actually inessential for this theorem; the theorem remains valid for any bounded closed set E (Section 2.2). For a proof, see Section 2.23. Again one has the corollary that a function f (x, y) that is continuous in a bounded closed region (or set) E is necessarily bounded on E ; that is, If (x, y)l 5 M for (x, y) in E and for a suitable choice of M. To determine the absolute maximum and minimum of a function f (x, y) defined on a bounded closed region E, one proceeds as for functions of one variable. One determines the critical points o f f inside E and the values o f f at the critical points; one then determines the maximum and minimum of f on the boundary. Among these values the desired maximum and minimum will be found. EXAMPLE Find the maximum and minimum of z = x2 x2 y2 5 1. One has

+

+ 2y2 - x on the set

Thus the critical point is at (i,O), at which z = -:. On the boundary of E, one has x 2 Y 2 = 1, SO that z = 2 - x - x2, -1 5 x 5 1. For this function we find the absolute maximum to be 2+, at the critical point x = -;; the absolute minimum is 0, at the end .x = 1. Thus the absolute maximum is 2 f , occurring at (- k$1, and the absolute minimum is - $, occurring at 0). Since the minimum occurs inside E, it is also a relative minimum, as can be checked by (2.146) through (2.149). One has here

+

i.

(i,

so that (2.147) is satisfied. The maximum and minimum on the boundary can also be analyzed by the method of Lagrange multipliers, to be explained in the following section. All of the foregoing extends to functions of three or more variables without essential modification. Thus at a critical point (x0, yo, zo) of w = f (x, y , z), all three derivatives awlax, aw/ay, a w l a z are 0, so that the directional derivative:

+

+

in the direction of an arbitrary unit vector u = cos cri cos B j cos y k is zero at the point (xo, yo, z o ) This directional derivative is the derivative d w l d s of a function of

153

154

Advanced Calculus, Fifth Edition

'

'

*

._

'

i l

r

7-.

one variable s , serving as coordinate on a line through (xo, yo, zo) in direction u. To analyze the critical point, one uses the second derivative d2w/ds2 = V U V w: u

a2w ax2

= -cos2a+2-

a2w a2w cos ol cos j3 + 2cos ol cos y (2.150) ax ay ax a 2

If this expression is positive for all u, then w has a relative minimum at (x0, yo, zo). Algebraic criteria for the positiveness of this "quadratic form7'can be obtained; see Section 2.21. - 4

Remarks. The tests for maxima or minima of f ( x ) given at the beginning of this section can be deduced from Taylor's formula with remainder (Section 6.17). There is a similar formula for functions of several variables (Section 6.21), and it can also be used to derive the tests for maxima and minima.

*2.20

EXTREMA FOR FUNCTIONS WITH SIDECONDITIONS LAGRANGE MULTIPLIERS

A problem of considerable importance for applications is that of maximizing or minimizing a function of several variables, where the variables are related by one or more equations, termed side conditions. Thus the problem of finding the radius of the largest sphere inscribable in the ellipsoid x 2+2y2 +3z2 = 6 is equivalent to minimizing the function w = x2 y2 z2, with the side condition x2 2y2 3z2 = 6. To handle such problems, one can, if possible, eliminate some of the variables by using the side conditions and eventually reduce the problem to an ordinary maximum and minimum problem such as that considered in the preceding section. This procedure is not always feasible, and the following procedure is often more convenient; it also treats the variables in a more symmetrical way, so that various simplifications may be possible. To illustrate the method, we consider the problem of maximizing w = f (x, y, z), where equations g(x, y, z ) = 0 and h ( x , y, z ) = 0 are given. The equations g = 0 and h = 0 describe two surfaces in space, and the problem is thus one of maximizing f(x, y, z) as ( x , y, z ) varies on the curve of intersection of these surfaces. At a maximum point the derivative o f f along the curve, that is, the directional derivative along the tangent to the curve, must be 0. This directional derivative is the component of the vector Vf along the tangent. It follows that Vf must lie in a plane normal to the curve at the point. This plane also contains the vectors Vg and V h (Section 2.13); that is, the vectors Vf,Vg,and V h are coplanar at the point. Hence (Section 1.3) there must exist scalars A, and h2 such that

+ +

+ +

Chapter 2 Differential Calculus of Functions of Several Variables at the critical point. This is equivalent to three scalar equations:

These three equations, together with the equations g(x, y , z ) = 0 , h ( x , y , z) = 0, serve as five equations in the five unknowns x , y , z , A,,A2. By solving them for x , y , z,one locates the critical points on the curve. It has been tacitly assumed here that the surfaces g = 0 , h = 0 do actually intersect in a curve and that V g and Vh are linearly independent. Cases in which these conditions fail are degenerate and require further investigation (cf. Section 2.22). The method described applies quite generally. To find the critical points of w = f (x,y , z , u , . . .), where the variables x , y , z , . . . are related by equations g(x, y, z , u , . . .) = 0 , h ( x , y , z , u , . . .) = 0 , . . . , one solves the system of equations

for the unknowns x , y , z , u , . . . , X I ,h 2 , . . . . The parameters A 12, . . . are known as Lagrange multipliers. To find the critical points of w = x y z , subject to the condition EXAMPLE x2 y2 + z2 = 1, one forms the function

+

f

+ hg = x y z + h(x2 + y2 + z2 - 1)

and then obtains four equations:

Multiplying the first three by x , y , z respectively, adding, and using the fourth equation, one finds h. = - i ( 3 x y z ) . Using this relation, one easily finds the 14 critical points (O,O, f I), ( 0 , f1 , O), ( f l , 0 , O), ( f f i / 3 , f&/3, f & / 3 ) . The first six listed are saddle points, whereas the remaining eight furnish four minima and four maxima, as simple considerations of signs show. The maxima and minima are absolute.

"-21

MAXIMA AND MINIMAOF QUADRATIC FORMS ON THE UNITSPHERE

One of the most important applications of the method of Lagrange multipliers is to the problem of maximizing or minimizing a quadratic form in n variables (Section 1.12): w = fc.1.

. . . . x., = ~ , ~ o , , x , x ,

(2.154)

155

156

Advanced Calculus, Fifth Edition subject to the side condition

We can interpret (2.155) as the equation of the unit sphere in En, so that our problem is that of maximizing or minimizing f on the unit sphere. As in Section 1.12, we let A = (aij) be the n x n matrix of coefficients of the quadratic form and can always assume that f is written in a form which makes A symmetric: aij = aji for all i and j, or A = A'. As in Section 1.12, we can then also write

f (xl, . . . , x,) = x'Ax,

(X = col (xl , . . . , x,)).

We proceed as in the preceding section and are led to the n + 1 equations:

Here f is as previously, and g(xl, . . . , x,) = 1 -x: - . . . -x:.

f (XI,. . . , x,)

4

Since A is symmetric,

+ + a12xzxl + al3xlx3 + a13x3xl+ . . . = allx: + 2a12xlx2 + 2a13xlx3+ . . . 2

= allxI

UIZXIXZ

(we are suggesting only the terms in xl), and accordingly,

Thus the Eqs. (2.156) become

Here the first n equations are equivalent to the equation A x = Ax.

Therefore they state that x (if nonzero) is to be an eigenvector of A, associated with the eigenvalue A (Section 1.11). The last equation states that 1x1 = 1, so that x is a unit vector and cannot be 0. Thus we are seeking the unit eigenvectors of A. If x is such an eigenvector, then f (xlr . . . ,x,) = x'Ax = x'hx = Ax'x = h(x:

+ .. . + x;)

@

= A.

Hence the eigenvalues of A provide the values of f at its critical points on the unit sphere. In particular, the absolute maximum o f f on the unit sphere equals the largest eigenvalue of A, and the absolute minimum off on the unit sphere equals the smallest eigenvalue of A. As in Section 1.13, all eigenvalues are real.

- *.'

-

Chapter 2 Differential Calculus of Functions of Several Variables EXAMPLE 1

so that A = [:

Let n = 2 and let

:].

Then the eigenvalues of A are the solutions of

Hence

The last form shows that the roots are always real (see Problem 12 following Section 1.13). If a = c and b = 0, the roots are equal and equal to (a c)/2. Otherwise, the larger root is obtained from the plus sign, the smaller from the minus sign. Also, if ac - b2 > 0, both roots have the same sign as a c. Thus if a c < 0, then both roots are negative. If ac - b2 < 0, the roots have opposite signs; if ac - b2 = 0, one root is 0, the other is a c. Since our side condition is x 2 y2 = 1, we can write x = cos a , y = sin a , and f becomes a cos2a 2b sin a cos a fc sin2a. When both roots are positive, f has a positive minimum and remains positive for all a ; when both roots are negative, f has a negative maximum and remains negative for all a . Thus the conditions found provide a new proof of the rules given in Section 2.19 for the function

+

+

+

+

+

+

In this example we have thus far considered only the nature and significance of the eigenvalues. To find the eigenvectors, and hence to find (x, y) that maximizes or minimizes f , one proceeds as in Section 1.11. For a general quadratic form f , one can imitate the example and ask for the conditions under which f has a positive minimum or a negative maximum on the unit sphere. We remark that every nonzero vector x = (XI,. . . , x,) can be written as k(u . . . , u,), where k = 1x1 and u = (u ,, . . . , u,) is a unit vector. Hence

,,

Accordingly, if f has a positive minimum on the unit sphere, then f is positive for all nonzero x. Conversely, if f is positive for all nonzero x, then f is positive on the unit sphere, in particular, and hence f has a positive minimum on the unit sphere. Similarly, f has a negative maximum on the unit sphere if and only if f is negative for all nonzero x. A quadratic form that is positive (negative) for all nonzero x is said to be positive dejnite (negative dejnite). Hence we can say, for example: f is positive dejnite i f and only i f all eigenvalues of A are positive. Another way to obtain these results is to choose new Cartesian coordinates (y,, . . . , y,) in En.As was pointed out in Section 1.13, for proper choice of the new

158

Advanced Calculus. Fifth Edition

~

coordinates, f becomes simply

where A,, . . . , A, are the (necessarily real) eigenvalues of A. It is thus clear that f is positive definite precisely when all the A's are positive. For further information on this topic, see Chapter 10 of Vol. 1 of the book by Gantmacher listed at the end of the chapter. For a general function F ( x l , . ..,x,), with critical point at (xf, . .. , x:), the method of Section 2.19 leads us to the quadratic form

where ( u l ,. . . , u,) is a unit vector and all derivatives are evaluated at the critical point. If this form is positive definite, then F has a minimum at the critical point. Hence ifall eigenvalues of the matrix (a2 ax; a x j ) arepositive, there is a minimum. Similarly, if all eigenvalues are negative, then there is a maximum. This matrix is called the Hessian matrix of F.

PROBLEMS

>

1. Locate the critical points of the following functions, classify them, and graph the functions: a) y = x3 - 3x, b)y=2sinx+sin2x, ~)y=e-~-e-~~. 2. Determine the nature of the critical point of y = xn (n = 2,3, . . .) at x = 0. 3. Determine the absolute maximum and absolute minimum, if they exist, of the following functions: a) y = c o s x , -$ 5 x 5 2 ' b) y = l o g x , O < x i l , c) y = tanhx, allx, d) y = + , allx. 1+x

4. Find the critical points of the following functions and test for maxima and minima: a) z = JF, b) z = 1 + x 2 + y 2 , c) z = 2x2 - xy - 3Y2- 3x 7y, , d) z = x2 - 5xy - y2, e) z = x2 - 2xy + y2, f ) z = x3 - 3xy2 + y3, g) z = x2 - 2x(sin y cosy) + 1, ' h ) z = ~ ~ ~ + x ~ ~ - x ~ , i) z = x3 Y3, j) z = x4 3x2y2 y4, , k) z = [x2 + (y + I ) ~ ] [ x ~(y - I ) ~ (interpret ] geometrically).

+

+

+ +

+

+

.

Chapter 2 Differential Calculus of Functions of Several Variables

5. Find the critical points of the following functions, classify, and graph the level curves of the functions: a)

z

=e-~Z-~2

b ) z = <4 - y4, c ) z = sin x cosh y, d ) z = *f.

6. Find the critical points of the following functions with given side conditions and test for maxima and minima: a) z = 3x 4 y , where x 2 y2 = 1, b ) z = x 2 + y 2 , where x4 y4 = 1, c ) z = x 2 + 24xy + 8y2, wherex2 y2 = 25, d ) w = x z, where x 2 y 2 z2 = 1,

+ +

+

+ +

+

+

7. Find the point of the curve

nearest to the origin (O,O, 0). 8. Find the absolute minimum and maximum, if they exist, of the following functions: a) z =

1 + X 2 + ~ f '

b) z = x y , x 2

all(x,y)

+ y2 I1,

c) w = x + y + z ,

x 2 + y 2 + z 2 5 1.

9. Determine whether the given quadratic form is positive definite:

+

+

a ) 3x2 2xy y2, b ) x 2 - x y - 2y2, 5 2 C) 7X1

+

+ 2x22 + 4 x 2 +~ ~'3x 23 '

4

T X I X ~

10. Prove the validity of the criterion (2.143) for a minimum, under the conditions stated. [Hint: The function VaVaf (xo, yo) is continuous in a for 0 5 CY 5 2 n and has a minimum M Iin this interval; by (2.143), M1> 0. By the Fundamental Lemma of Section 2.6, a z l a x and az/ay have differentials at (xo, yo). Show that this implies that

+

+

where x = xo s cos a, y = yo s sin or (s > 0 ) and Ic I can be made as small as desired by choosing s sufficiently small. Choose S so that 1c 1 < for 0 < s < 6 and show that

4~~

Accordingly, f increases steadily, as one recedes from (xo, yo) on a straight line in the neighborhood of radius S of (no, yo).]

159

I

160

Advanced Calculus, Fifth Edition 11. (The method of least squares) Let five numbers el, e2, e3, e4, e5 be given. It is in general impossible to find a quadratic expression f (x) = ax2 bx c such that f (-2) = el, f (- 1) = e2, f (0) = e3, f ( 1 ) = e4, f (2) = e5. However, one can try to make the "total square-error"

+ +

+

+

+

+

E = (f(-2) - ell2 (f(-1) - e212 (f (0)- e312 (f (1) - e412 (f (2) - es)' as small as possible. Determine the values of a , b, and c that make E a minimum. This is the method of least squares, which is basic in the theory of statistics and in curve-fitting (see Chapter 7). 12. Find the shortest distance d between the two skew lines x=1+2t,

y =2-3t,

z=5t

and x = 2 + r ,

y=3+2r,

z=1-r

by minimizing the squared distance function for variable points on the two lines. 13. Let f be a positive definite quadratic form aijxix, in (xl , . . . , x,) (a;, = aji). Show that each level set E : f (xl, . . . , x,) = c (c > 0) is a bounded closed set [Hint: Let ho > 0 be the smallest eigenvalue of A = (ai,) and show that f (x) = c implies 1x1' 5 ho/c. This shows that E is bounded. To show E is closed, proceed as in Problem 7 following Section 2.4 to show that if xo is not in E, so that f(f (xo)- c) > 0, then some neighborhood of xo has no point in E.] 14. Let A and B be symmetric n x n matrices and let f (xl , . . . , x,) = x' Ax, g(xl , . . . , x n ) = x'Bx be the corresponding quadratic forms. Let g be positive definite. Use Lagrange multipliers to seek critical points of f on the set g ( x l , . . . , x,) = 1 (a bounded closed set by Problem 13). a) Show that, at each critical point x, Ax - hBx = 0 and x'Bx = 1, where A is a root of the nth degree equation det (A - hB) = 0,

(*)

and that f (x) = A. b) Show that if h l , h2 are distinct roots of (*) and X I , x;? are corresponding choices of x, then x', Bxz = 0. (Compare Problem 15 following Section 1.13.) 15. Show that Problem 14 can be reduced to the problem of Section 2.21 by a suitable linear transformation x = Ez, where E is a nonsingular n x n matrix. [Hint: First set x = Cy, where C is an orthogonal matrix such that C'BC = BI = diag(wl, . . . , w,), with all w, > 0, so that as in Section 1.13. Then set y = Dz, where D = diag(l/pl'2,

. . . , 1/wL'2), SO that

Verify that x satisfies the critical point conditions Ax - hBx = 0, x'Bx = 1, if and only if z satisfies the conditions A2z - hz = 0,

Z'Z

= 1,

where A2 = D'AI D , A1 = C'AC, so that A2 is symmetric, and that these conditions are those for a critical point of f2(z) = z1A2z,with A2 symmetric, on the unit sphere. Finally remark that x = Ez, where E = C D and E is nonsingular.] Remark A further orthogonal transformation z = Cl w diagonalizes A2. The combined nonsingular transformation x = F w , with F = C D C l , diagonalizes both coefficient matrices A , B of the given quadratic forms.

161

Chapter 2 Differential Calculus of Functions of Several Variables

Figure 2.22

Level curves o f f (x, y) with Vf # 0.

Throughout this chapter the condition that a derivative or Jacobian is zero has played an important role. Thus in Section 2.10 the condition that the Jacobian not be zero was needed to obtain the derivatives of implicit functions; in Section 2.19 the condition that all partial derivatives be zero was used to locate critical points. In this section we consider these questions from a more general point of view, with emphasis on certain extreme cases that are of importance. Let w = f ( x , y ) be given in a domain D. If V w = (af / a x ) i ( af /ay)j is not 0 in D, then the level curves f ( x , y ) = const of w are well-defined-curves, one through each point of D. This follows from the implicit function theorem of Section 2.1 1; at each point ( x , ,y l ) , either af / a x # 0 or af l a y # 0, so that (2.70) gives a definite derivative for the implicit function. The family of level curves is thus without singularity, as shown in Fig. 2.22. Singularities will be introduced if V w = 0 at certain points of D. These are precisely the critical points considered in Section 2.19. Figure 2.21 illustrates some of the possible complications. The extreme case is that in which V w = 0 in D, that is, every point of D is a 0 and af l a y 0 in D , and one concludes that f is critical point. Here af / a x constant in D:

+

-

k

THEOREM

Let f ( x , y ) be defined in domain D and let

in D. Then there is a constant c such that

Proof. Let PI: ( x y ) and P2:( x 2 ,y2)be two points of D such that the line segment PI Pz lies in D. Let P : (.u, y ) vary on this segment a n s t s be the distance P I P . The directional derivative of f at P in the direction PIP2 is then equal to d f l d s ; since af / a x = 0, af l a y = 0, one has d f Ids = 0. Hence by the familiar theorem for functions of one variable, f , as a function of s , is constant on PI P2. Thus f ( X Iyl) , = f ( x Z ,y Z ) = c for some c. Every point of D can be joined to PI by a

1

162

Advanced Calculus, Fifth Edition broken line (cf. Section 2.2 and Fig. 2.1); by repetition of the argument just given, one concludes that f (x, y) = f (xl , yl) = c for every point (x, y) of D. The theorem is now proved. Now let two functions u = f (x, y), v = g(x, y) be given in D. Suppose that Vf # 0 in D and V g # 0 in D, so that both functions have well-defined level curves, as in Fig. 2.22. "In general" the two families of level curves determine curvilinear coordinates in D and a mapping of D onto a domain in the uv-plane; this will be the case if the Jacobian a( f , g)/a(x, y) f 0 in D, as a study of the theorem on implicit functions shows. It should be noted that the condition a( f ,g)/a(x, y) = 0 is the same as the condition Vf x V g = 0, that is, that Vf and V g are collinear vectors. For

When these vectors are collinear, the level curves f = const and g = const are tangent, the curvilinear coordinates are interfered with, and the mapping to the uv-plane may degenerate. The extreme case here is that for which Vf x V g = 0 in D or, equivalently, a( f , g)/a(x, y) = 0 in D. In this case, each level curve off is a level curve of g and vice versa. For a tangent vector to a level curve f = const is a vector perpendicular to the normal vector (af /ax)i + (af /ay)j; thus T = -(a f /ay)i (af /ax)j is such a tangent vector, as shown in Fig. 2.23. The component of V g in the direction of this tangent is 0, since

+

and the Jacobian is 0 by assumption. Thus the directional derivative of g along the curve is 0, and g must be constant. Therefore ifa( f , g)/a(x, y) 0 in D, then the level curves off and g coincide. Conversely, iff and g have the same level curves, then a( f , g)/a(x, y) 0 in D, for the argument just given can be reversed.

Figure 2.23

Tangent to level curves.

Chapter 2 Differential Calculus of Functions of Several Variables

EXAMPLE 1 Let f (x, y) = ex sin y and g ( x , y) = x and0 < y > n.Then 1

cot y

+ log sin y for 0 < x < 1

=eXcosy -eXcosy =0.

In this example the functions f and g are related by an identity:

in the domain considered; that is, g is simply the function log f , a "function of the function." From this fact it is clear that on a level curve f = const = c , one must also have g = const = log c. In general, two functions f and g related by an identity

in a given domain D, are termed functionally dependent in D. Here F[u, v] is a function of the variables, u, v such that F [f ( x , y), g(x, y)] is defined in D, and in order to rule out degenerate cases we assume V F # 0 for the range of u , v involved. THEOREM If f ( x , y) and g(x, y) are differentiable in the domain D and are functionally dependent in D, then

so that the level curves off and g coincide. Conversely, if (2.161) holds and Vf # 0, V g # 0, then, in some neighborhood of each point of D, f and g are functionally dependent.

ProoJ: Let f and g be f u n c v l l y dependent in D, so that (2.160) holds for a suitable F [ u , v ] . On differenaating (2.160) with respect to x and y and using the chain rules, one obtains the identities:

Since a F / a u and a F / a v are not both zero, these equations are consistent only if the "determinant of the coefficients" is zero (Section I S ) , that is, only if

Thus (2.161) holds. Conversely, let (2.161) hold, so that f and g have the same level curves (Fig. 2.24). The equations

then define a mapping from thexy-plane to the uv-plane. This mapping is degenerate, for along each level curve of f and g, u and v have constant values, so that the entire level curve maps into a single point in the uv-plane. If one considers a particular

163

164

$

4

Advanced Calculus, Fifth Edition

Figure 2.24

9

Level curves of functionally dependent functions.

point ( x l ,y l ) of D , then on proceeding from this point in a direction normal to the level curve, f must either increase or decrease, since V f # 0; similarly, g must either increase or decrease. Thus a sufficiently small neighborhood in the xy-plane is mapped by (2.164) on a curve in the uv-plane expressible either as u = $ ( v ) or as v = $(u). Thus f ( x , y ) - $ ( g ( x , y ) ) = 0 in the neighborhood. Along the normal, in terms of a suitable coordinate t , f and g are differentiable functions of t and at ( x , , y,) u l ( t ) = af/ a n = Vf . n # 0 in Eq. (2.1 17) and similarly vl(t) # 0, so that d u / d v = u l ( t ) / v ' ( t )= +'(v). Hence $ is differentiable, as is F(u, v ) = u - @ ( v ) and f , g are functionally dependent in the neighborhood. The proof just given brings out the significance of the condition a ( f , g)/ a ( x , y ) E 0 for the mapping (2.164): This mapping maps D not onto a domain but onto a curve or several curves. For the functions f = ex sin y , g = x +log sin y the corresponding curve is given by part of the graph of log u - v = 0. The results obtained can be generalized to the case of three functions of three variables or, in general, to n functions of n variables. Thus for three functions of three variables, functional dependence

is equivalent as previously to the condition

9

This in turn is equivalent to the statement that the three vectors Of, V g , V h are coplanar at each point, so that the three families of level surfaces have a common tangent direction at each point. One can also consider the case of n functions of m variables: f l ( x , , . . . , x,,,), . . . , f n ( x l , . . . , x,) and allow for several functional dependencies among them.

+

Linear case. Here f,( x , , . . .) = a,~ x , . . . + a,,x, for i = 1, . . . , n or, in vector notation, f(x) = Ax, with A = ( a , , ) a constant n x m matrix. We restrict attention to linear dependencies, of form Z I fi

+ . . . + znfn

0,

where z l , . . . , z, are constants, not all 0. (A general functional dependence F ( f i , . . . , f,) r 0 implies a linear one; see Problem 11.) Equation (2.165) is

3

"

Chapter 2 Differential Calculus of Functions of Several Variables equivalent to z'Ax = 0 for all x, where z = col ( z l , . . . , z,), and hence to z ' A = 0 or (2.166)

A'z = 0,

If A has rank r , then as in Section 1.17 the solutions z of (2.166) form a subspace W of dimension n - r of V n .We choose a basis z l , . . . , zn-, for W and obtain n - r independent linear dependencies: where z , = col ( z , , , . . . , zan),and all other linear dependencies are linear combinations of these. If n 5 m and A has maximal rank n, then there are no such relations and the n functions fi are functionally independent. If m < n, then r 5 m < n and the dependencies exist. In all cases A maps V m onto an r-dimensional subspace of V n and the mapping is one-to-one precisely when r = m. General case. The results for the linear case extend in a natural way to the general case of n functions f , ( x l , . . . , x m ) (i = 1 , . . . , n), defined and differentiable in domain D.Functional dependence of f l , . . . , f, in D is understood to mean the existence of h functions F , ( u l , . . . , u , ) ( j = 1 , . . . , h ) , defined and differentiable in a domain D, of E n , such that the composite functions F J ( f ~ ( x. .~.),, . . . , f n b i , . . .I),

J = 1,.

..,h ,

(2.168)

are defined in D and are identically 0 in D.Here it is assumed that the Jacobian matrix (aF, / a u k ) has maximum rank h in D,,so that h 5 n. When these conditions hold, differentiation yields the equations

We denote the Jacobian matrix ( a f k / a x e )by A. Equation (2.169) then asserts that A'z, = 0 at each point x of D ,where A is evaluated at x and z[, is the jth row of ( a F , / a u k ) ,evaluated at the corresponding point u of D,.Since the last matrix is assumed to have rank h , z l , . . . , zh are linearly independent; accordingly, h 5 n - r , where r is the rank of A . Therefore, r 5 n - h : THEOREM

Functional dependence: F , ( f ~ ( x l.,. . , xm), . . ., fn(x1, . . . xm)) = 0 in D 9

(2.170)

for j = 1, . . . , h, implies that the Jacobian matrix A = ( a f ,l a x ; ) has rank r 5 n - h at each point of D. To obtain a converse, we assume that A has constant rank r < n in D.Then one can prove that, in an appropriate neighborhood Dl of each point x0 of D ,the functions f l , . . . , fn are functionally dependent, so that they satisfy relations of form (2.170) in D l with h = n - r, where F l , . . . , Fh are as above. In fact, one can show that, if r 5 n, then in terms of appropriate curvilinear coordinates 61, . . . , in D l and q l , . . . , qn in a neighborhood of the corresponding point u0 = f(xO),the mapping u = f ( x ) becomes simply the linear mapping

em

165

166

~ns-

Advanced Calculus, Fifth Edition

-9

In the original coordinates r], = r],(ul, . . . , u,) and the functional dependencies are as in (2.170), with F I ( U I.,. . , u,) = v,.+I,. . . , F ~ ( u I. , . , Un) = qn(ul, . . . , u,). If r = n , there are no functional dependencies. For proofs of the results just stated one is referred to pp. 86-87 of Differential Geometry, Lie Groups, and Symmetric Spaces, by S. Helgason (American Mathematical Society, Providence, R.I., 2001). The range of the mapping u = f(x) just considered is often a smooth curve, surface or hypersurface in En.As the expression of the mapping in local coordinates shows, S has dimension r. One calls the mapping a submersion of D onto S.

.

+ + +

EXAMPLE 2 Let fl(x1, x2, x3) = f2(x1, XZ, x3) = sin (XI xz x 3), f3fx1,x2, x3) = cos (XI x2 ~ 3 )Then . A = (af,/ax,) has rank 1 in E~ and there are two dependencies FJ(f l , f2, f3) E 0; one choice with (a Fj/auk) of rank 2 is

+ +

(See Problem 12 below.)

PROBLEMS 1. A function f (x, y), defined for all (x, y), satisfies the conditions

f(x,O)=sinx,

af

- -= 0.

a~

Evaluate f (~r/2,2),f (n, 3), f (x, 1). 2. Two functions f (x, y) and g(x, y) are such that

Vf = Vg in a domain D.Show that f =g+c for some constant c . 3. Determine all functions f (x, y) whose second partial derivatives are identically 0. 4. A function f (x, y), defined for all (x, y), is such that = 0. Show that there is a function g(x) such that

f (x, Y ) = g(x). 5. Determine all functions f (x, y) such that

&&= 0 for all (x, y). [Hint: Cf. Problem 4.1

6. Show that the following sets of functions are functionally dependent: a) f =:, g = x5+2y ;

+ +

d

4

b) f = x2 + 2xy + y2 2x 2y, g = exeY; c) f = X ~ ~ - X ~ ~ + Xg = ~ xZ y, + x - y + z , ~ = X ~ + ~ ~ + Z ~ - ~ ~ Z d) f = u + v - x , g = x - y + u , h = u - 2 ~ + 5 ~ - 3 y . 7. Find an identity relating each of the sets of functions of Problem 6. 8. Plot the level curves of the functions f and g of Problem 6(a) and (b).

Chapter 2 Differential Calculus of Functions of Several Variables 9. Let f ( x , y ) and g ( x , y , u) be such that

when u = f ( x , y ) . Then show that f ( x , y ) and g [ x , v , f ( x . y)l are functionally dependent. 10. Let u ( x , y ) and v ( x , y ) be harmonic in a domain D and have no critical points in D. Show that if u and v are functionally dependent, then they are "linearly" dependent: u = av + b for suitable constants a and b . [Hint: Assume a relation of form u = f ( v ) and take the Laplacian of both sides.] 11. a) Show that if linear functions u = a x + b y , v = c x + d y are functionally dependent then they are linearly dependent: hu(x, y ) + k v ( x , y ) _= 0 for suitable constants h and k. [Hint: Differentiate the identity F ( a x + b y , c x + d y ) = 0 with respect to x and y .] b) Generalize the result of part (a) to n functions of m variables. 12. With reference to Example 2 in Section 2.22, verify that A has rank 1 in E 3 , that the F, can be chosen as stated, that the range of u = f ( x ) is an ellipse and that (aF,/auk)has rank 2 in a domain D, containing the ellipse. (Here u = f ( x ) is a submersion of E3 onto the ellipse.)

*2.23 REALVARIABLE THEORY THEOREM ON MAXIMUM AND MINIMUM In this section we prove some basic theorems on sets and on functions of real variables. As will be seen, the concept of sequence is important for the development. (By "sequence" we here mean "infinite sequence.") We take the real number system itself as known and think of real numbers as those representable as decimal expressions such as 2.3175, -1.3333. . . , and 3.14159.. . . These include the integers 0, k 1 , f2, . . . , and every real number x satisfies k 5 x < k + 1 for a unique integer k. They also include the rational numbers: numbers representable as plq, where p and q are integers with q # 0. If we represent the rational numbers as points on a line, the number axis, then these points do not fill the line: there are infinitely many points left out, such as those representing the real numbers 2/2, rr and e and, in general, all the irrational numbers (those which are not rational). For each irrational number x and each positive integer q, we can find a unique integer k such that k/q < x < (k l)/q. We need only choose k as above so that k < q x < k 1 (equality is excluded here); thus x is approximated by rational numbers above and below within distance r = 1/q. The sequence s, is said to converge to the real numbers (written s, + s ) or have limit s if for each r > 0 there is an integer N for which Is, - s I < r for n > N. The limit, when it exists, is unique, since if also s, + s f ,then

+

+

for n sufficiently large; this holds for every r > 0, so that s' = s . A sequence s, is said to be rnonotone increasing if s, _< s,+l for all n and monotone decreasing if s, 2 s,+, for all n; if the inequality is strict (< or >), then

167

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Advanced Calculus, Fifth Edition S2

SI I

c

-

$3 1

-

-

-

-

1

s

I I I I

I I

* T kL

I

B I

h

10

Figure 2.25

Convergence of bounded monotone sequence.

these are said to be strictly increasing or strictly decreasing. In all cases the sequence is called monotone. A sequence s, is said to be bounded above if for all n , s, 5 B for some number B and bounded below if for all n , s, > A for some number A; it is said to be bounded if it is bounded above and bounded below. We observe that a monotone increasing (decreasing) sequence bounded above (below) is necessarily bounded. THEOREM A

3

Eve'ry bounded monotone sequence converges.

4

%is

Proof. Let the sequence be s,, n = 1 , 2 , . . . and let it be increasing, with s, 5 B for all n. We then try to find s by locating it to the nearest integer below s , then to the nearest tenth and so on, as in Fig. 2.25. In detail we let h ( B c h 1 , c Isl < c + 1 for integers c, h , so that necessarily c 5 h . Also c ( s, < h + 1 for all n. We consider the finite set of integers c, c + 1 , . . . , h and let p be the largest of these that is attained or exceeded by some s,. Thus p 5 s, < p 1 for n sufficiently large. We consider the finite set of numbers

+

+

+

and let p (kl 110) be the largest number in this set attained or exceeded by some s,, so that kl is one of the integers 0, 1. . . . , 9 , and

for n sufficiently large. Continuing in this way, we obtain a sequence k l , kz,. . . , k,, . . .of integers, each having one of the values 0 , 1, . . . , 9 . Lets be the real number p 0.klk2 . . . k, . Then from our construction we have s, ( s for all n; for if s, > s , then we would have failed to choose one of the k, as large as possible. Also, for each m,

+

a .

S,

2p

+ O.klk2 . - km > s -

for all n sufficiently large. Given r > 0, we choose m so large that lo-" < r ( m can be obtained from the decimal expression for 6 ) . Thus for n sufficiently large, s - r < • s, 5 s, so that Is, - sl < r. Accordingly, s, + s . A set E of real numbers is said to be bounded above if there is a number B such that x ( B for all numbers x in E. We call B an upper bound of E . For example, if E is the set of numbers

1 2

n

2 ' ~ ' . . .n3+ l " . . '

--

drl, r

i

i2

Chapter 2 Differential Calculus of Functions of Several Variables then E is bounded above, and B = 1 or any number larger than 1 is an upper bound of E. Similarly, E is bounded below if x 2 B for all x in E, for some number B ; B is then a lower bound of E . THEOREM B Let E be a set of real numbers that is nonempty and bounded above. Then E has a least upper bound; that is, there is a number Bo such that Bo is an upper bound of E and Bo 5 B for every upper bound of E. Proo$ We can imitate the proof of Theorem A. There are integers that are not upper bounds of E-for we can choose an integer less than some x in E (E is nonempty). There is a largest such integer, sl.For E has an upper bound B , and every integer that is not an upper bound must be less than B. Similarly, there is a largest integer kl,0 5 k l 5 9, such that s2 = s1 (kl/lO) is not an upper bound of E. Continuing, we obtain a monotone increasing sequence s, such that

+ R

and s, is not an upper bound of E but s, that for every x in E, for all n ,

+ (11 lo"-')

is an upper bound. It follows

Now the sequence s, is bounded above (by B , as earlier). Therefore by Theorem A, s, converges to some Bo. By (2.17 I),

so that Bo is an upper bound of E. On the other hand, s, < B for all n implies that Bo 5 B. Therefore Bo is the least upper bound (and is clearly unique).

CORO1,LARY TO THEOREM B If a set E of real numbers is bounded below, then E has a greatest lower bound Ao. The proof is left as an exercise (Problem 1 following). One abbreviates the least upper bound of E and the greatest lower bound of E by lub E,

glb E.

A sequence t, is said to be a subsequence of s, if tl = s,, , t 2 =s,,, . . . , tk = s,, , . . . , where n < n2 < . . < nk c nk+~< . . . . We remark that if the sequence s, converges, then the subsequence t, also converges, with the same limit (Problem 2). THEOREM C (Weierstrass-Bolzano Theorem) Lets, be a bounded sequence of real numbers. Then s, has a convergent subsequence. Cr Proof. Let E be the set of all real numbers x such that s, >_ x for infinitely many values of n . Since the sequences, satisfies A 5 s, 5 B for all n,for some A and B , E

Advanced Calculus, Fifth Edition

I.(!

contains at least the number A, but E contains no number larger than B. Therefore E is bounded above. Let so = lub E. Then any numbers less than so is in E. Otherwise, s, 2 s for only finitely many values of n, and hence no number x of E could be greater than s. That is, s would be an upper bound of E less than so, and that is impossible. Also, any number s greater than so is not in E, since so = lub E. Now we consider the intervals

For each k, by the two cases just considered, s, > so - 2-k for infinitely many values of n, but s, 2 so 2-k for only a finite set of values of n. Therefore at least one of the s,, s,, , can be chosen to lie in the interval, and nk can be made as large as desired. In particular, we can successively choose s,, , s,?, . . . SO that

+

and nl < n2 < . . < nk < . . . . Clearly, s,, is the desired subsequence, converging to so as k + oo. We now consider sequences of points in E N . It will be convenient to represent each point P = (xl, . . . , XN)by the vector x = (xl, . . . , xN) of V N . AS in Section 2.4, we write d ( P , Q) for the distance between P and Q, so that d ( P , Q ) = Ix - yl if x, y are the corresponding vectors. As in Section 2.4, we have the rules I.

(i) d(P, Q) 2 0, d(P, Q) = 0 only for P = Q, (ii) d(P, Q >= d(Q, P>,

.-

"

+

(iii) d(P, S) 5 d(P, Q> d(Q, S>. Furthermore, as in Problem 10 following Section 2.4, if x = (x,, . . . , xN),then (iv) 1x1 < c implies 1x11 < e , . . . , IXNI < e, (v) /xll < S, . . . , IxNI < S together imply 1x1 < NS.

A sequence P, of points in E~ is said to converge and have limit Po if ' lim d(P,, Po) = 0.

n+03

We write P, + Po. As before, the limit is unique. For N = 1, this is consistent with the definition given earlier of convergence of sequences of real numbers. THEOREM D Let P, = (xr, x,", . . . , x i ) be a sequence of points in EN.Then P, converges to Po = (xf, x i , . . . , x i ) if and only if

Chapter 2 Differential Calculus of Functions of Several Variables The proof, which is similar to that of Problem 10(c)and (d) following Section 2.4, is left as an exercise (Problem 3). A set G in E N is said to be bounded if 1x1 ( K for some constant K, for all x = (xl, . . . , x,) in G. This is again consistent with the definition for N = 1 of bounded sets of real numbers. THEOREM E (Weierstrass-BolzanoTheorem for EN) Let P, be a sequence of points in the bounded set G of E N .Then Pn has a convergent subsequence.

Proof. For simplicity we take N = 2, so that P, = (x: , x,") = xn. Since Ixn1 5 K, we conclude from Rule (iv) that Ix; I ( K, 1x2"I 5 K for all n. Thus x; is a bounded sequence of real numbers. By Theorem C it has a convergent subsequence x;\ The sequence

is a subsequence of Pn, and x,"' is a bounded sequence of real numbers. By Theorem C again, this sequence has a convergent subsequence. By relabeling we can again denote this by x,"" it is still a subsequence of x i , and the corresponding sequence x;"~ a subsequence of a convergent sequence and therefore also converges (Problem 2 below). Therefore by Theorem D the new sequence P,, = (x;" xx,"") converges and is the desired subsequence of Pn. THEOREM F A set G in E~ is closed if and only if, for each convergent sequence P,, in G, the limit of the sequence is in G.

For the proof, see Problem 12 following Section 2.4. We consider a mapping f from a set G in E N into E M .This is said to be continuous at Po in G if for each E > 0, there is a 6 > 0 such that d( f (P), f (Po)) < r whenever d ( P , Po) < 6. The mapping f is said to be continuous in G if it is continuous at every point of G. THEOREM G Let f be a mapping of a set G in E N into E ~Then . f is continuous at Po in G if and only if f (P,) + f (Po) for every sequence Pn in G converging to Po.

Proof. Let f be continuous at Po and let Pn + Po, with PI, P2, . . . all in G . Given E > 0, we choose 6 > 0 as in the definition of continuity above. Then we choose n, sothatd(P,, Po) < S forn > n,.Thend(f(Pn), f(Po)) < E forn > n,, so that f (Pn) + f (PO). Conversely, let f be such that f (P,) + f (Po) wherever P, + Po. Suppose f is not continuous at Po. Then for some E > 0 for every 6 > 0 there is some P in G such that d(P, Po) < 6 but d( f (P), f (Po)) > c. We take 6 successively equal to 1/2,1/4, . . . , 1/2", . . . to obtain a sequence Pn such that d(P,, Po) < 2-" and d( f (P,), f (Po)) > E. Then clearly, P, + Po,but f (P,) does not converge to f (Po), contrary to assumption. Hence f must be continuous at Po. THEOREM H Let G be a bounded closed set in E N . Let f be a continuous mapping of G into EM.Then the range of f is also bounded and closed.

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dQ

5

z

4

4

Proof. We observe first that to say that G is bounded is equivalent to requiring that d ( 0 , P) 5 K , for some constant K , for all P in G; here 0 is the origin of E N . If the range of f is not bounded, then we can find a sequence P, in G such that d ( O 1 ,f (P,)) > n for n = 1 , 2 , . . . ,; here 0' is the origin of E M .By Theorem E, P, has a convergent subsequence P,, , P,, + Po. By Theorem F, Po is in G. By TheoremG, f(P,,) + f (Po)ask + oo.Butnk < d ( O 1 ,f (P,,)) 5 d ( O 1 ,f ( P o ) ) + d( f (Po),f (P,, )). As k + oo, the last term has limit 0, but nk -+ oo.This would imply oo _< d ( O 1 ,f (Po)),which is impossible. Hence the range o f f is bounded. By Theorem F we show that the range is closed by showing that if Q, = f (P,) is a convergent sequence in the range, then its limit is also in the range. As earlier, we choose a subsequence P,, converging to Po in G and then f (P,, ) + f ( Po) by continuity. But the sequences f (P,) and f (P,,) have the same limit. Therefore Q,, -+ f (Po),which is in the range. Therefore the range is closed.

C

THEOREM I Let E be a nonempty, bounded closed set of real numbers. Then lub E and glb E belong to E . ProoJ

Let xo = lub E. For each k = 1 , 2 , . . . there must be some xk in E such that

Otherwise, xo could not be the least upper bound of E. Clearly, xk + xo as k + GO. Since E is closed, xo is in E , as was asserted. The proof is similar for glb E.

THEOREM J Let G be a bounded closed set in E N and let f be a continuous real-valued function on G, hence mapping G into E ' . Then f takes on an absolute maximum and minimum on G.

'

ProoJ By Theorem H the range of f is a bounded, closed set F of real numbers. By Theorem I, F contains a = glb F and b = lub F. Therefore a 5 f ( P ) 5 b for all P in G and f ( P I )= a for at least one point P I , and f ( P 2 )= b for at least one point P2. Thus f takes on its absolute minimum, a , at PI and its absolute maximum, b, at P2.

This theorem includes as special cases the theorems stated in Section 2.19: That

a continuous function f ( x ) on the interval a 5 x 5 b and a continuous function f ( x , y) on a bounded closed region in the plane take on their absolute minimum and maximum. Remark.

A bounded closed set in E N is also called a compact set in E~

PROBLEMS .

1. Prove the corollary to Theorem B, 2. Prove: If s, is a convergent sequence of real numbers with limit c, then every subsequence of the sequence s, converges to c. 3. Prove Theorem D.

1

Chapter 2

Differential Calculus of Functions of Several Variables

4. Prove the following: a) A closed interval a 5 x 5 b is a bounded closed set. b) If v { ( x )and y2(x) are continuous for a 5 x 5 b and y l ( x ) 5 y2(x) for a 5 x 5 b, then the set G: a 5 x 5 b , y l ( x ) c y 5 y2(x)is a bounded closed set. Is G a closed 5.

6.

7.

8.

region? Explain. Prove the Intermediate Value theorem for functions of one variable: I f f ( x )is continuous for a 5 x 5 b and f ( a ) < 0, f ( b ) > 0, then there exists an xo, a ixo < 6 , such that f (yo)= 0. [Hint: Let E be the set of x on the interval for which f ( x ) < 0 and let xo be lub E.] Prove: If f is a real-valued function defined on E ~then , f is continuous on E~ if and only if for every open interval a < x 6 b the set of all P for which a < f ( P )< b is open. Let G be a closed set in E ~Let. Q be a point of E N not in G and let f ( P )be the distance from an arbitrary point P of G to the point Q. a) Show that f is continuous on G. b) Show that f has a positive minimum at a point Po of G. Is the point Po unique? Let G and H be two closed sets in E N without common points. Show by example that the greatest lower bound (glb) of the distance between a point of G and a point of H can be 0, but if G is bounded the glb is positive and is attained for some pair of points.

Suggested References Apostol, Tom M., Mathematical Analysis: A Modern Approach to Advanced Calculus, 2nd ed. Reading, Mass.: Addison-Wesley, 1974. Buck, R. C., Advanced Calculus. 3rd ed. New York: McGraw-Hill, 1978. Coddington, Earl A., and Norman Levinson, Theory of Ordinary Differential Equations. New York: McGraw-Hill, 1955. Courant, Richard J., and Fritz John, Introduction to Calculus and Analysis, vol. 1, New York: Interscience, 1965 and vol. 2, New York: John Wiley & Sons, 1965. Gantmacher, F. R., Theory oj'Matrices, trans]. by K. A. Hirsch, 2 vols. New York: Chelsea Publishing Co., 1960. Kaplan, Wilfred, Maxima and Minima. New York: John Wiley & Sons, 1999. Rudin, W., Principles of Mathematics Analysis. 3rd ed. New York: McGraw-Hill, 1976. Struik, Dirk J., Lectures on Classical Differential Geometry, 2nd ed. Reading, Mass.: Addison-Wesley, 1961.

Vector Differential Calculus

3.1 INTRODUCTION +

8lt5a

In Section 2.13 the path of a moving point P is described by giving its position vector r = ?$ as a function of time r . It is shown how this vector function can be differentiated to give the velocity vector of the moving point. This operation can properly be termed a part of vector differential calculus. In this chapter there is again a natural physical model-a fluid in motion. At each point of the fluid, one has a velocity vector v, the velocity of the "fluid particle" located at the given point. Such vectors are defined for all points of the fluid and together form what is termed a vectorjield. This is illustrated in Fig. 3.1. The field may change with time or remain the same (stationary flow). One can again trace the paths of the individual particles, the "stream lines," and determine the velocity vector v = d r l d t for each. However, one can also consider the velocity field at a given time as describing a vector v, which is a function of x , y , and z , that is, of position in space. The vector function v(x, y , z ) can then be differentiated with respect to x , y, and z ; that is, one can consider the rate and manner in which v varies from point to point in space. The description of the variation of v turns out to require not merely partial derivatives, but special combinations of these, the divergence and curl. At each point of the field a scalar, div v, the divergence of v, and a vector, curl v, will be defined. The divergence measures the net rate at which matter is being transported away from the neighborhood of each point, and the condition div v

E

0

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Advanced Calculus, Fifth Edition

Figure 3.1

Vector field in space.

1 .-

Figure 3.2

I

Vector field in the plane.

describes the incompressible flow of a fluid. The curl is essentially a measure of the angular velocity of the motion; in the case when the fluid is rotating as a rigid body with angular velocity w about the z axis, the curl of v is everywhere equal to 2wk (Problem 16 following Section 3.6). There are other important physical examples of vector fields, such as force fields arising from gravitational attraction or from electromagnetic sources; the familiar experiment showing the effect of a magnet on iron filings illustrates the latter. In many cases the vectors of the field are parallel to a fixed plane and form the same pattern in each plane parallel to this plane. In this case the study of the field can be reduced to a 2-dimensional problem. One is thus led to study vectorBelds in the plane, as illustrated in Fig. 3.2.

3.2 VECTORFIELDSAND SCALAR FIELDS If to each point (x , y , z) of a domain D in space a vector v = v(x , y , z) is assigned, then a vectorjeld is said to be given in D. Each vector v of the field can be regarded as a "bound vector" attached to the corresponding point ( x , y , z). If v is expressed v,j v,k, then these components will also vary in terms of components, v = v,i from point to point, so that one has

+ +

v = v d x ,Y , z)i

Tt

i

+ v,(x, Y , z)j + v,(x, y , z)k,

(3.1)

-

Chapter 3 Vector Differential Calculus

Figure 3.3

Example 1.

Figure 3.4

Example 2.

~IW 5

that is, each vector field is equivalent to a triple of scalar functions of the three variables x , y , z . d

EXAMPLE 1 Let v = xi + yj + zk.Here v = 0 at the origin; at other points, v is a vector pointing away from the origin, as illustrated in Fig. 3.3. EXAMPLE 2 Let v = -yi + xj. Here the vectors can be interpreted as the velocity vectors of a rigid rotation about the z axis. This is illustrated in Fig. 3.4. EXAMPLE 3 Let F denote the gravitational force exerted on a particle of mass m at P: (x, y, z) by a mass M concentrated at the origin. Newton's law of gravitation gives

Mmr F = -k--, r2 r where r =

177

a

*-

I

3 and k is a universal constant. Here r/r is a unit vector, so F has

magnitude

and the force is inversely proportional to the square of the distance. The notion of vector field can be specialized to two dimensions. Thus a vector field v in a domain D of the xy-plane is given by

v = v,(x, y)i

+ v,(x, y)j,

(3.3)

where v,(x, y) and v,(x, y) are two scalar functions of x and y defined in D. Twodimensional fields arise in applications mainly in connection with planarproblems, that is, problems concerning a vector field v in space such that v is always parallel to the xy-plane and v is independent of z; that is, v, = 0 and v, and v, depend only on x and y, as in (3.3). From (3.3),one can construct the vectors v first in the xy-plane and then, using exactly the same vectors, in any plane parallel to the xy-plane. i /

EXAMPLE 4

Let F be the field

This is illustrated in Fig. 3.5 and can be interpreted as the electric force field due to two infinite straight wires, perpendicular to the xy-plane at (1, 0) and (- 1, 0), homogeneously and oppositely charged with electricity. a

Y

Figure 3.5

Example 4.

179

Chapter 3 Vector Differential Calculus If we assign to each point of a domain D in space a scalar, rather than a vector, we obtain a scalarBeld in D. For example, the temperature at each point in a room defines a scalar field. If coordinates x , y, z are introduced, the scalar field determines a function f ( x , y , z ) in D. One can also consider scalar fields in the plane; each such scalar field is described by a function f ( x , y ) . It will be seen that scalar fields give rise to vector fields (for example, the field " of the vector grad f )and that vector fields give rise to scalar fields (for example, the field of the scalar Ivl). %

Let a scalar field f be given in space and a coordinate system chosen so that f = f ( x , y , z ) and is defined and differentiable in a certain domain of space; thus the first partial derivatives of f exist in this domain. They form the components of'the vector grad f , the gradient of the scalar f . Thus one has

af grad f = -i ax

'J

af + -jafay + -k. a2

For example, if f = x2 y - z 2 , then grad f = 2 x y i

+ x2j - 2 z k .

.f i : '

.*

Formula (3.5) can be written in the following symbolic form:

i

f

where the suggested multiplication actually leads to a differentiation. The expression in parentheses is denoted by the symbol V and is called "del" or "nabla." Thus

V is a "vector differential operator." By itself, the V has no numerical significance; it takes on such significance when it is applied to a function, that is, in forming

The operator V will be shown to be exceedingly useful. It was shown in Section 2.14 that the directional derivative of the scalar f in the direction of the unit vector u = cos ai cos Bj cos y k is given by

+

+

This shows that grad f has a meaning independent of the coordinate system chosen:' Its component in a given direction represents the rate of change o f f in that direction. In particular, grad f points in the direction of maximum increase of f .

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Advanced Calculus, Fifth Edition

4i

The gradient obeys the following laws: grad(f

+ g) = grad f + grad g,

grad(fg) = f grad g

(3.10)

+ g grad f ;

(3.11)

that is, with the V symbol,

These hold, provided grad f and grad g exist in the domain considered. The proofs are left for the problems. If f is a constant c, (3.1 1 ) reduces to the simpler condition: grad (cg) = c grad g

(c = const).

(3.13)

If the terms in z are dropped, the preceding discussion specializes at once to two dimensions. Thus for f = f ( x , y), one has grad f

af. + -J, af. = V f = --I ax

a

v = -axi +

ay

(3.14)

a.

-J.

ay

C

4

PROBLEMS 1. Sketch the following vector fields:

a) v = (x2 - y2)i+ 2xyj, c) v = -yi xj k,

+ +

+

b) u = (X- y)i+(x y)j, d) v = -xi - yj - zk.

e

g

2. Sketch the level curves or surfaces of the following scalar fields: 3. Determine grad f for the scalar fields of Problem 2 and sketch several of the corresponding vectors. 4. Show that the gravitational field (3.2) is the gradient of the scalar

5. Show that the force field (3.4) is the gradient of the scalar

6. Prove (3.10) and (3.11). 7. Prove: If f (x, y, z) is a composite function F(u), where u = g(x, y, z), then grad f = F1(u)grad g. V 8. Prove: grad = [g grad f f f grad g].

6 $

9. If f = f (xl, . . . ,x,), then the Hessian matrix of f is the matrix

.

.1

Chapter 3 Vector Differential Calculus a) Find this matrix for the functions w = x 3 y

+

2 x 2 ~ 4x22 ~

+ x2xg + 5 ~ ~ +x x3x2 1 + 2x32.

- y 3 z and w = x:

+ 2x1x2 + 5x1x3 +

b) Show, under appropriate assumptions, that H is symmetric. c) If z = f ( x , y), show that V,Vgz = [cos ,b sin ,b] H[cos a sina]' (see Section 2.21).

3.4 THEDIVERGENCE OF A VECTORFIELD Given a vector field v in a domain D of space, one has (for a given coordinate system) three scalar functions v, , v,, v,. If these possess first partial derivatives in D, we can form in all nine partial derivatives, which we arrange to form a matrix:

From three of these the scalar div v, the divergence of v, is constructed by the formula:

It will be noted that the derivatives used form a diagonal (principal diagonal) of the matrix and the divergence is the trace (Problem 6 following Section 1.11). For example, if v = x2i - xyj xyzk,then

+

Formula (3.15) can be written in the symbolic form: div v = V v;

(3.16)

for, treating V as a vector, one has

~he.definitionof the divergence at first appears to be quite arbitrary and to depend on the choice of axes in space. It will be seen in Section 3.8 that this is not the case. The divergence has in fact a definite physical significance. In fluid dynamics it appears as a measure of the rate of decrease of density at a point. More precisely, let u = u(x, y, z, t) denote the velocity vector of a fluid motion and let p = p(x, y, z,t) denote the density. Then v = pu is a vector whose divergence satisfies the equation div v = --.a~ at

181

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Advanced Calculus, Fifth Edition

I

This is in fact the "continuity equation" of fluid mechanics. If the fluid is incompressible, this reduces to the simpler equation div u = 0.

(3.18)

The law (3.17) will be established in Chapter 5; the derivation of (3.18) from (3.17) is considered in Problem 2 following Section 3.6. The divergence also plays an important part in the theory of electromagnetic fields. Here the divergence of the electric force vector E satisfies the equation div E = 4np,

(3.19)

where p is the charge density. Thus where there is no charge, one has div E = 0.

(3.20)

The divergence has the basic properties:

+ div v, div (fu) = f div u + grad f

/

div (U+ V) = div u

..$.

'

(3.21) (3.22)

u;

that is, with the nabla symbol:

The proofs are left to the problems.

From the remaining six partial derivatives of the square array of Section 3.4, one constructs a new vector field, curl v, by the definition:

It will be noted that each component is formed of elements symmetrically placed relative to the principal diagonal. The curl can be expressed in terms of V, for one has

curlv=Vxv=

I I1 a

.

(3.24)

The determinant must be expanded by minors of the first row, that is, so as to yield (3.23). The fact that this vector field has a meaning independent of the choice of axes will also be shown in Section 3.8. The curl is important in the analysis of the velocity fields of fluid dynamics and in the analysis of electromagnetic force fields. The curl

,

- A

i f

Chapter 3 Vector Differential Calculus can be interpreted as measuring angular motion of a fluid (see Problem 16 following Section 3.6) and the condition curl v = 0

(3.25)

for a velocity field v characterizes what are termed irrotationalJows. The analogous equation curl E = 0

(3.26)

for the electric force vector E holds when only electrostatic forces are present. The curl satisfies the basic laws:

+ v) = curl u + curl v, curl (f u) = f curl u + grad f x u.

curl (u

(3.27) (3.28)

The proofs are left to the problems.

As a result of the new definitions, we now have at our disposal the operations listed in Table 3.1. The theory of vector algebra, discussed in Section 1.2, concerns the properties of the algebraic operations and their combinations. The theory of vector differential calculus concerns the theory of the differential operations (g) to (k) and their combinations with each other and with the algebraic operations (a) to (f). The combinations of (g) and (h) with the algebraic operations are discussed in Problem 3 following Section 2.13. The combinations of (i), (j), and (k) with (a) and (c) are discussed in Sections 3.3, 3.4, and 3.5. The results can be summarized in the one rule: operator on sum = sum of operators on terms.

(3.29)

TABLE 3.1 Operation

Algebraic Operations

Differential Operations

Symbols

I

(a) sum of scalars (b) product of scalars (c) sum of vectors (d) scalar times vector (e) scalar product (f) vector product ( (g) derivative of scalar (h) derivative of vector (i) gradient of scalar (j) divergence of vector (k) curl of vector

I

I

Vf = grad f V.v=divv V x v 5 curl v

Where Discussed (Section No.)

183

Advanced Calculus, Fifth Edition The combinations of (i), (j), (k) with (b) and (d) are also discussed in Sections 3.3 to 3.5. The results include the important case of scalar constant times scalar or vector. Here we have the general rule: operator on scalar constantfactor = scalarfactor times operator;

(3.30)

that is, a scalar constant can be factored out. Thus V(cf) = c V f , V . (cu) = c V . u, and so on. The rules (3.29) and (3.30) characterize what are called linear operators; thus grad, div, and curl are linear operators. If one considers the other possible combinations, one obtains a long list of identities, some of which will be considered here. The proofs are left to the problems. All derivatives occurring are to be assumed continuous. Curl of a gradient.

Here one has the rule: curl grad f = 0.

(3.31)

This relation is suggested by the fact that curl grad f = V x (Vf ), that is, has the appearanceof the vector product of collinear vectors. There is an important converse: if curl v = 0, then v = grad f for some f ;

(3.32)

further assumptions are needed here, and the rule (3.32) must be used with caution. A proof and full discussion are given in Chapter 5. A vector field v such that curl v = 0 is often termed irrotational. Divergence of a curl. Here one concludes that

r u.. 1

div curl v = 0;

(3.33)

this relation is again suggested by a vector identity, for div curl v = V . (V x v), so that one has an expression resembling a scalar triple product of coplanar vectors (Section 1.4).Again there is a converse:

. .

if div u = 0, then u = curl v for some v;

(3.34)

as with (3.32) there are restrictions on the use of (3.34), and one is again referred to Chapter 5. A vector field u such that div u = 0 is often termed solenoidal. Divergence of a vector product.

div (U x Divergence of a gradient.

Here one has V)

= v - curl u - u . curl v.

(3.35)

If one expands in terms of components, one finds that

divgrad f =

a2f a2f -+-+-. ax2 ay*

a2f a22

3

(3.36)

The expression on the right-hand side is known as the Laplacian of f and is also denoted by Af or by f , since div grad f = V . ( Vf ). A function f (having continuous second derivatives) such that div grad f = 0 in a domain is called harmonic

v2

2

&

185

Chapter 3 Vector Differential Calculus

,

in that domain. The equation satisfied by f :

is called Laplace's equation (see Sections 2.15 and 2.17).

dS

Here an expansion into components yields the relation:

Curl of a curl.

curl curl u = grad div u - ( v 2 u , i + v 2 u , j

+v2u,k).

(3.38)

If one defines the Laplacian of a vector u to be the vector

then (3.38) becomes curl curl u = grad div u - v 2 u .

(3.40)

This identity can be written as an expression for the gradient of a divergence: grad div u = curl curl u

+v2u.

(3.41)

The identities listed here, together with those previously obtained, cover all of interest except for those for gradient of a scalarproduct and curl of a vectorproduct; these two are considered in Problems 13 and 14.

PROBLEMS 1. Prove (3.21) and (3.22). -EIB 2. Prove that the continuity equation (3.17)can be written in the form t

wm ..

or, in terns of the Stokes derivative (Problem 12 following Section 2.8), thus: DP + p div u = 0. Dt Prove that (3.17) reduces to (3.18) when p = const. It will be shown in Chapter 5 that the same simplification can be made when p is variable, provided that the fluid is incompressible. This follows from the fact that DplDt measures the variation in density at a point moving with the fluid; for an incompressible fluid this local density cannot vary. 3. Prove (3.27)and (3.28). 4. Prove (3.31).Verify by applying to f = 5. For the given vector field v, verify that curl v = 0 and find all functions f such that grad f = v. a) v = 2xyzi x2zj x2yk. b) v = eXy[(2y2 yz2)i (2xy + xz2 2)j 2zkl. 6. Prove (3.33).Verify by applying to v = x2yzi - x3y3j xyz2k.

*-

+

+

+

+

+ +

+

186

Advanced Calculus, Fifth Edition

7. a) Given the vector field v = 2xi + y j - 3zk, verify that div v = 0.Find all vectors u such that curl u = v. [Hint: First remark that on the basis of (3.32), all solutions of the equation curl u = v are given by u = uo + grad f , where f is an arbitrary scalar and uo is any one vector whose curl is v. To find uo, assume uo . k = 0.1 b) PIroceed as in (a) for v = yi z j xk. 8. Provc:(3.36). Verify that the function f of Problem 4 is harmonic in space (except at the origil-I). [This function, which represents the electrostatic potential of a charge of +1 at the 01rigin, is in a sense the fundamental harmonic function in space, for every harmonic function in space can be represented as a sum, or limit of a sum, of such functions.] 9. Prow:(3.35). 10. Provc:(3.38). 11. Provc:the following identities: a) diIV[Ux (V x w)]= ( u . w)divv-(u.v)divw+grad(u. w ) . v - g r a d ( u . v ) . w , b) diIV (grad f x f grad g) = 0, c) curl (curl v grad f ) = curl curl V, d) v2f = div (curl v grad f ). These should be established by means of the identities already found in this chapter and not by expanding into components. 12. One defines the scalar product u . V , with u on the left of the operator V, as the operator

+ +

+

4

+

This is thus quite unrelated to V - u = div u. The operator u . V can be applied to a scalar f :

thus an associative law holds. The operator u . V can also be applied to a vector v:

where the partial derivatives & / a x , one has

. . . are defined just as is d r l d t in Section 2.13; thus

a) Show that if u is a unit vector, then (u . V ) f = V, f . b) Evaluate [(i - j) . V ] f . c) Evaluate [(xi - yj) . v ] ( x 2 i - y2j z2k). Prove the identity (cf. Problem 12): grad (u . v) = (u . V ) V (v . V ) u (u x curl v) (V x curl u). Prove the identity (cf. Problem 12): curl (u x v) = u div v v div u + (v . V ) u - (u . V)V. Let n be the unit outer normal vector to the sphere x 2 + y2 + z2 = 9 and let u be the vector (x2 - z2)(i - j + 3k). Evaluate alan (div u) at (2,2, 1). A rigid body is rotating about the z-axis with angular velocity w. Show that a typical particle of the body follows a path

+

13. 14. 15. 16.

+

+

+

-

= r cos (wt

+ a)i + r sin (wt + a)j + zk,

# i

1

Chapter 3 Vector Differential Calculus where a,r, and z are constant, and that at each instant the velocity is v=ox3,

where o = wk (the angular velocity vector of the motion). Evaluate div v and curl v. 17. A steady fluid motion has velocity u = yi. Show that all points that move do so on straight lines and that the flow is incompressible. Determine the volume occupied at time t = 1 by the points which at time t = 0 fill the cube bounded by the coordinate planes andtheplanesx = 1 , y = 1,z = 1. 18. A steady fluid motion has velocity u = xi. Show that all points either do not move or else move on straight lines. Determine the volume occupied at time t = 1 by the points which at time t = 0 fill the cube of Problem 17. [Hint: Show that the paths of the individual points are given by x = c l e f , y = cz, z = c3, where c f ,cz, c3 are constants.] Is the flow incompressible?

The discussion of the preceding sections has been confined to a fixed rectangular coordinate system in space. We now consider the extension of the theory to curvilinear coordinates, such as cylindrical or spherical coordinates. New coordinates u, v, w can be introduced in space by equations x = f (u,v , w),

y = g ( u , v, w ) ,

z

= h(u, v, w ) .

(3.42)

It will be assumed that f , g, and h are defined and have continuous first partial derivatives in a domain D lof uvw-space and that Equations (3.42) can be solved uniquely for u, v, and w as similar functions of x, y , z:

The inverse functions will be defined in a domain D of xyz-space, and we term u, v , w curvilinear coordinates in D. We assume that the Jacobian

is positive in D l . If v and w are assigned constant values vo and wo while u is allowed to vary, Eqs. (3.42) define a curve in D, on which u is a parameter. As the numbers vo and wo are varied, one obtains a family of curves in D, like the parallels to one of the axes in rectangular coordinates. For example, in spherical coordinates p , $ , 0 , the curves 4 = $o, 0 = Oo ( p variable) are rays through the origin. If we write r = xi yj zk for the position vector of a point (x, y, z), the Eqs. (3.42) can be interpreted as defining a vector function r = r(u, v, w). When v = vo, w = w o ,this is the vector representation r = r(u, vo, w o )of the curve of the preceding paragraph. The tangent vector to this curve is defined as in Section 2.13 to be the derivative of r with respect to the parameter u. Here we must write a r / a u for the derivative, to indicate that v and w are held constant. Similarly, a r / a is ~ tangent to a curve u = const, w = const, and a r / a w is tangent to a curve u = const, v = const. This is illustrated in Fig. 3.6.

+ +

187

188

Advanced Calculus, Fifth Edition ar -

aw

Figure 3.6

Curvilinear coordinates in space.

We write:

Accordingly,

is the "speed," in terms of time u, with which a curve v = vo, w = wo is traced, and ds = a! du is the element of distance. , ar/a w and Jacobian J can be expressed in The tangent vectors a r / a ~ar/av, terms of x, y, and z by the equations

ar ar ar - = J ( V G x V H ) , - - J ( V H x V F ) , - = J ( V F x VG), (3.46) au av aw

The gradients V F , V G , V H can be expressed in terms of u, v , w by the equations

The proofs are left to Problems 1 to 3 following Section 3.8. Because of the assumption J > 0, the vectors arlau, ar/a v , ar/a w form a positive triple; because of (3.47), VF, V G , V H also form a positive triple. We note two further sets of identities: 1 ar 1 ar 1 ar -- = curl(GVH), -- - curl(HVF), ---- curl (FVG); (3.49) J au J av J aw

The first reduces to (3.46) on application of the identity (3.28). The second then follows from (3.33).

9

4

Chapter 3 Vector Differential Calculus The curvilinear coordinate system defined by (3.42) and (3.43) is said to be orthogonal if the tangent vectors ar/au, a r / a v , a r / a w at each point of D form a triple of mutually perpendicular vectors. It will be seen that when this is the case, important simplifications in the formulas are possible. The most commonly used curvilinear coordinates are orthogonal, and for this reason we shall confine attention to this case. In this and the following section the coordinates will be assumed to be orthogonal. As a first consequenceof orthogonality,we remark that ( l l a )arlau, (118)ar/av, ( l l y )a r / a w are a positive triple of mutually perpendicular unit vectors. Hence

Accordingly, by (3.44,

A similar reasoning applies to V G and V H ; we conclude that a V F, B V G , y V H are also mutually perpendicular unit vectors and

The surfaces F = const, G = const, H = const must hence meet at right angles; they form what is called a triply orthogonal family of surfaces. Conversely, when the vectors V F, V G , V H are mutually perpendicular throughout D, the coordinates must be orthogonal (Problem 4 following Section 3.8). A curve in D can be described by equations: x = x(t), y = y ( t ) , z = z ( t ) , or, by (3.43), in terms of the curvilinear coordinates by the equations u = u:t), v = v(t),w = w ( t ) .The element of arc d s on such a curve is defined by the equation

Hence

Since the coordinates are orthogonal, we conclude:

189

190

Advanced Calculus, Fifth Edition Now a du ,9, d v , y d w are the elements of arc on the u curves, v curves, and w curves, respectively. The expression (3.54) is built up as a sum of squares of the elements in three coordinate directions just as in (3.53).This is one of the basic properties of orthogonal coordinates. The above derivation shows that (3.54)can hold only when arlau, a r l a v , a r / a w are mutually perpendicular, so that (3.54) itself can be used to define what is meant by orthogonal coordinates. We note also that a,9y du d v d w can be interpreted as the volume dV of an "elementary rectangular parallelepiped." Hence by (3.44) and (3.5l ) ,

d V = a b y d u d v d w = J d u d v d w = a(x' " du d v d w . a(u, v , W ) This formula will be discussed in Chapter 4 . From Eqs. (3.52)and the identity (3.31)we deduce the important rule: curl

ar (- -) au 1

~2

= 0,

curl

(--)avar 1

82

= 0,

curl

1 ar (? ),,

= 0.

(3.55)

Now let a vector field p be given in D. The vector p can be described by its components p,, p,, p, in terms of the given rectangular system. However, at each point of D, the vectors ( 1 1 ~a~r /)a u , (1/,9) a r / a v , ( l l y )a r l a w are a triple of mutually perpendicular unit vectors. Hence we can write

in terms of the components p,, p,, p, in the directions of the three unit vectors. It should be emphasized that the triple of unit vectors in general variesfiom point to point in D . By (3.52)we can also write

The components p, , p, , p, can be computed from the components p, , p, , p, in the rectangular system. For example,

Similarly, the components p, , p,, p, can be computed from pu , p, , p,:

Chapter 3 Vector Differential Calculus If the representation (3.56') is used, one finds:

Since the vectors ( l / a ) a r l a u , ( l / p ) a r l a v , ( l / y ) a r l a w form apositivetriple of mutually perpendicular unit vectors, the operations scalar times vector, sum of vectors, scalar product, and vector product can be carried out in terms of components in the directions of these unit vectors just as in terms of x , y , and z components. In particular,

On the other hand, because the base vectors vary from point to point, the differential operations become more complicated: 1 a4

1 34 [grad 41, = - -

[grad 41. = - -, au

B av

+

a

[grad#],=--

7

1

,

(3.60)

Y aw

div p = -[ a ( ~ ~ ~ G u )( ~ ~ ~%(uBpw) a" ) , ffBy au

+

To prove (3.60), we use (3.57):

(

a4 ay [grad 41, = - -- +-+-a ax a~ a~ a~ a~ a~ 1

a4 ax

(3.61)

=-a

a~

The other components are found in the same way. We remark that [grad $1. is the directional derivative of 4 along a u curve, that is, d+/ds in terms of arc length s on the curve. Since ds = a du, the result (3.60) follows at once. To prove (3.61), we use (3.51) and (3.56) to write:

Advanced Calculus, Fifth Edition

:s

The divergence of p is the sum of the divergences of the terms on the right-hand side. By (3.22) and (3.50) the divergence of the first term is

By (3.60) this can be written as

a

(grad Byp,). J

1 ar a 1 a = - [grad Byp,l, = --(Byp,) a a~ J J a~ --

1 a = -- (BYP~). ~ B au Y

This gives the first term on the right-hand side of (3.61); the others are found in the same way. The proof of (3.62) is left to Problem 5 following. From (3.60) and (3.61) we obtain an expression for the Laplacian in orthogonal curvilinear coordinates:

Remark. If the curvilinear coordinates are not orthogonal, the very notion of vector components must be generalized. This leads to tensor analysis (see Section 3.9).

The formulas (3.60), (3.61), (3.62) can be applied to the special case in which the new coordinates are obtained simply by choosing new rectangular coordinates u, v, w in space. This can be accomplished by choosing a positive triple i l ,j l , kl of unit vectors and then choosing axes u , v, w through an origin 0 ,: (XI,yl , zl) having the directions il ,jl , kl , respectively (Problems 13 and 14 following Section 1.15). The new coordinates (u, v, w ) of a point P : ( x , y , z) are defined by the equation:

One then finds

similar expressions are found for v and w . Also, 1

and similar expressions are found for y and z. These two sets of equations correspond to (3.43) and (3.42). We note that in all cases the functions involved are linear. The quantities a, B, y can be evaluated for this case without computation. For, since the new coordinates are rectangular and no change of scale is made, one must have

for element of arc on a general curve. Hence

rl

Chapter 3 Vector Differential Calculus If we substitute these values in (3.60),(3.61),(3.62),we find that the basic formulas ( 3 3 , (3.15),(3.23)reappear, with x replaced by u , y by v , and z by w . Foi-example,

This shows that the fundamental deJinitions ( 3 . 9 , (3.15), (3.23) of grad, div, curl are not dependent on the particular coordinate system chosen. If a change of scale is made, the formulas will be altered. Physically, this corresponds to a change in unit of length (for example, from inches to feet), and one could hardly expect a temperature gradient, for example, to have the same value in degrees per inch as in degrees per foot. For this reason, in practice one must always specify the units used. It has been assumed throughout that the Jacobian J is positive; this implies, in particular, that in the case just considered the vectors i,, j l , kl form a positive triple. If J is negative, one finds that the only change in (3.60), (3.61), (3.62) is a reversal of sign in the components of the curl. In particular, if new rectangular coordinates are chosen, based on a negative triple i l ,j l , k l , the curl as defined by (3.23) relative to these axes is the negative of the curl relative to the original x , y, and z axes. If the orientation of the m e s is reversed, the curl reverses direction. This is not surprising if one thinks of the curl as an angular velocity vector. On the other hand, the gradient vector and divergence do not depend on the orientation. The vector product components in (3.59) also change sign if the orientation is reversed. This could be predicted from the very definition of the vector product in Section 1.2. For further discussion of change of coordinates in space, see Chapter 4 of Classical Mechanics, 3rd Edition, by H. Goldstein, C. P. Poole, and J. L. Safko (Boston, MA: Addison-Wesley, 2002). If the Jacobian J is 0at a point P, it is in general impossible to use the curvilinear coordinates at P; in particular, the vectors a r l a u , a r / a v , a r / a w are coplanar, so that one cannot express an arbitrary vector p in terms of components p,, p,, p,. Furthermore, the point P will in general correspond to many values of the coordinates (u, v , w ) ; that is, the inverse functions (3.43)become ambiguous at P. However, let us assume that the functions (3.42)remain continuous and differentiable in the domain Dl of the coordinates (u, v , w ) and have values in a domain D; then the chain rules (2.35) imply that every function U(x, y , z ) that is differentiable in D becomes a function

U [ f ( u ,v, w ) , g(u, v , w ) , h(u, v , w ) ] that is defined and differentiable in Dl. Accordingly, even though the inverse transformation (3.43) may fail to exist, scalar functions of ( x , y, z ) can be transformed into functions of ( u , v , w ) without difficulty. Precisely this situation arises in cylindrical and spherical coordinates. For example, in cylindrical coordinates the Eqs. (3.42) are as follows:

these functions are defined and differentiable as often as desired for all real values of r, 0, z ; as ( r , 8 , z ) ranges over all possible combinations, ( x , y , z ) ranges over all

193

rT

I

I

1

I

194

Advanced Calculus, Fifth Edition

.

of space. An inverse transformation:

can be defined by suitably restricting 8; but in no way can this be defined as a triple of continuous functions when ( x , y, z) ranges over a domain D including points of the z axis. It is precisely when r = 0 that the Jacobian J = r (Problem 6) is zero. It is shown in Problem 6 that the Laplacian in cylindrical coordinates has the expression:

This expression becomes meaningless when r = 0. However, if we know that U , when expressed in rectangular coordinates, has continuous first and second derivatives on the z axis, then V 2 u must also be a continuous function of (r, 0, z ) for r = 0. Under these assumptions, V'U can be obtained from (3.66) for r = 0 by a limit process. For example, if U = x2

+ xZY= r 2cos2@ + r 3cos2e sine,

then

(3.66) gives the indeterminate expression

While this function is indeterminate for r = 0, it has a definite limit as r + 0; the limit is 2 and is independent of 8. By cancelling r 2 from numerator and denominator we automatically remove the indeterminacy and assign the correct limiting value for r = 0:

A similar discussion applies to spherical coordinates p , 4,O. In Problem 7 it is shown that J = p2 sin 4, so that J = 0 on the z axis. The Laplacian is found to be

This is indeterminate on the z axis (p = 0 or sin 4 = 0); when V'U is known to be continuous on the z axis, the values of V'U on this line can be found from (3.67) by a limit process. It is also possible to obtain the values of the Laplacian at the troublesome points directly in terms of derivatives. For example, V'U can be computed at the origin (p 0) by the formula: =>

9

4

Chapter 3 Vector Differential Calculus The three terms on the right are simply the three terms of the Laplacian

at the origin.

PROBLEMS 1. Prove (3.48). [Hint: Use the result of Problem 5 following Section 2.12.1 2. Prove (3.47). [Hint: Use (3.48) and the rules (1.19) of Section 1.2.1 3. Prove (3.46). [Hint: Use (3.48) and the rules (1.19) of Section 1.2.1 4. Prove that when the vectors V F, V G , V H are mutually perpendicular in D, the coordinates are orthogonal. 5. Prove (3.62). [Hint: Use (3.56) to write: 1 ar) 1 ar) 1 ar) p = ( u p u ) -- + B P ~ -- +(YP",) -- a 2 a~ p2 a~ y2 a~ t Use (3.28) and (3.55) to show that

(

(

(

+ g r a d ( B ~ u )X

curl p = grad (up,) x

-p(2). av ....

and use (3.59) and To compute [curl p],, take the scalar product of both sides with (3.60) to compute the scalar triple products.] 6. Verify the following relations for cylindrical coordinates u = r, v = 0, w = z: a) the surfaces r = const, 0 = const, z = const form a triply orthogonal family, and

b) the element of arc length is given by ds2 = d r 2

+ r2de2 + dz2;

c) the components of a vector p are given by p, = p x c o s O + p y s i n O ,

pe = - p x s i n O + p y c o s e ,

Pz=Pz;

au fi ;?B, au au d) grad U has components: T,

x;

+ %+r t ] ;

r) ciiv p = f [$(rp,) f ) curl p has components:

g) V ~ U is given by (3.66). 7. Verify the following relations for spherical coordinates u = p, v = 4 , w = 8: a) the surfaces p = const, 4 =const, 0 = const form a triply orthogonal family, and a(x, Y , Z) J = = p2 sin 4 ; a(p, 6 0 ) b) the element of arc length is given by ds2 = dp2 p2&p2 p2 sin2 I$ do2;

+

+

195

1

196

Advanced Calculus. Fifth Edition c) the components of a vector p are given by p, = p , s i n ~ c o s 8 + p y s i n ~ s i n 8 + p , c o s ~ , p@= p x c o s ~ c o s 8 + p y c o s ~ s i n-8p,sin$,

+ p, cos8; au r a,u , 1 au ; d) grad U has components v, 1 a e) div p = - F - - [ S ~ ~ ~ ~+(P&(P$ P ~ Psin41 ~ ) + p$@]; P sin@ pe = -px sin8

f ) curl p has components:

g) V ~ is Ugiven by (3.67).

8. (Curvilinear coordinates on a surface) I

x = f ( u , v),

Equations

Y = g(u, v),

z = h(u, V)

can be interpreted as parametric equations of a surface S in space. They can be considered as a special case of (3.42), in which w is restricted to a constant value, while (u, v) varies over a domain D of the uv plane; the surface S then corresponds to a surface w = const for (3.42). We consider u, v as curvilinear coordinates on S. The two sets of curves u = const and v = const on S form families like the parallels to the axes in the xy-plane. Graph the surface and the lines u = const, v = const, for the following cases: a) sphere: x = sin u cos v, y = sin u sin v, z = cos u; b) cylinder: x = cos u, y = sin u, z = v; c) cone: x = sinh u sin v, y = sinh u cos v , z = sinh u. 9. For the surface of Problem 8, assume that f , g, h have continuous first partial derivatives in D and that the Jacobian matrix

has rank 2 in D (Section 1.17). Show that each point (uo, vo) of D has a neighborhood Do in which the correspondence between points (u, v) and points of the surface is oneto-one. [Hint: If, for example, a( f, g)/a(u, v) # 0 at (uo, vo), then apply the Implicit Function Theorem of Section 2.10, as in Section 2.12, to obtain inverse functions u = 4(x, y), v = $(x, y ) of x = f (u, v), y = g(u, v) in Do. Show that each point (x, y, z) of the surface, for (u, v) in Do, comes from a unique (u, v) in Do.] 10. Let a surface S be given as in Problems 8 and 9 and let r = x i yj zk. a) Show that a r l a u and a r l a v are vectors tangent to the lines v = const, u = const on the surface. b) Show that the curves v = const, u = const intersect at right angles, so that the coordinates are orthogonal, if and only if

+ +

Chapter 3 Vector Differential Calculus C)

Show that the element of arc on a curve u = u ( t ) ,v = v ( t ) on S is given by ds2 = E du2 2 F d u d v G d v 2 ,

+

+

-F = -ar. - =ar- - ax ax +--+-ay ay az az

a~ a~

a~ a~

a~ a~

a~ a ~ '

d) Show that the coordinates are orthogonal precisely when ds2 = E du2 G dv2. e) Let u = u ( t ) ,v = v ( t ) and u = U ( t ) ,v = V ( r )be two curves on S meeting at a point Po of S for t = to, t = to, so that u(to) = uo = U(ro),v(to) = vo = V(ro). Show that the angle 8 between the corresponding velocity vectors d r l d t at to, d r l d r at TO (assumed to be nonzero vectors) satisfies

+

where E, F , G are evaluated at (uo,vo) and u' = ul(to),v' = vl(to),U' = U 1 ( t O ) , V' = V1(t0). f ) Show that if the paths of part (e) are coordinate lines:

+

+

u ( t ) = uo t - to, v ( t ) = vo, U ( r ) = uo, V ( r )= vo r - TO. then cos6 = F(EG)-'I2 at (uo,vo). g) Show from the hypotheses of Problem 9 that ar/au and ar/av are linearly independent in D. h) Show that E > 0, G > 0, E G - F2 > 0. [Hint: E > 0,G > 0] follow from (g). By Problem 12 (a) following Section 1.5,

EG - F~ = larlau x ar/av12; show from (g) that this is > 0.1 i) Show from the meaning of ds2 in terms of arc length that E du2 2F d u d v G d v 2 is a positive definite quadratic form and hence E > 0,G 0,E G - F~ > 0. [Hint: See Example 1 in Section 2.21.1 For further theory of surfaces, see the book by Struik listed at the end of this chapter.

+

+

When nonorthogonal curvilinear coordinates are introduced, the methods of the preceding section are no longer adequate for the analysis of the basic vector operations. The desired analysis can be carried out with the aid of tensors, which we proceed to develop briefly here. We first consider the very simple case of a change of scale in space. Let ( x , y, z) be given Cartesian coordinates and let (Z, j , Z ) be new coordinates, where

197

198

Advanced Calculus, Fifth Edition A being a positive constant scalar. Thus we have changed scale in the ratio A : 1, and l/A is our new unit of distance. For a point moving on a path we have hitherto assigned a velocity vector v = (v, , v,, v,) and thought of this as a definite geometric object, represented by a directed line segment. But the components of v, as a velocity vector, depend on the coordinate system; that is, here, where we are considering a change of scale, these components depend on the unit of distance chosen. In the (x, y, Z) coordinates we would assign dx y =vy = ddt ' dt ' but in the ( f , J , Z) coordinates we would assign Ux

By virtue of Eqs. (3.69), d f l d t = A dxldt, so that

Uz

d2 dt '

=-

r s

ex = Av,,

and in general,

@,

Thus in the new coordinates (f , J , Z) we assign to v the new components (Ex, , U,), which are h times the previous components. Now a vector v can also be obtained as the gradient vector of a function f : v = grad f , so that

af

af

af , v y = - , v, = vz=-. ax ay a2 If we change scale by (3.69), we would still like v to be the gradient o f f . But in the new coordinates, f becomes f (f/A, Y/A, .?/A) = f ( f , Y, Z), and this has gradient

Hence now

Thus when we change scale by (3.69), the three components of the gradient vector are divided by A. This result is to be expected, since the gradient vector measures the rate of change of f with respect to distance (in various directions), and we are changing the unit of distance. If for example, A = 2, then the new unit of distance is half the old one, and the amount of change in f per unit of distance is half as much as before, whereas velocity components are twice as large as before, since one covers twice as many units of distance per unit of time. The two different rules (3.70) and (3.7 1) show that we really have two different ways of assigning components to vectors. In the case of (3.70), one is dealing with a "contravariant vector," and in the case of (3.71), one is dealing with a "covariant vector." In obtaining the two types of components we have emphasized the change in unit of distance. However, in changing coordinates we can always keep in mind the original unit of distance (as a standard of reference), so that all distances can ultimately be restated in the original units. This concept of a standard unit of distance will be important in the development to follow.

Chapter 3 Vector Differential Calculus In tensor analysis, one'allows changes of coordinates that are much more general than changes of scale. It is more convenient to number our coordinates as (x' , x2,x3) and simply to refer to the (x') coordinates. We can in fact reason generally about n-dimensional space and allow i to go from 1 to n. The use of superscripts rather than subscripts is required for a certain consistency of all the tensor notations. If we introduce new coordinates (xi), then we have equations relating new and old coordinates. We consider these equations only in a neighborhood D of a certain point and assume that they define a one-to-one mapping with inverse XI r. i'= f i ( x l , ..., xn), i = 1, ..., n, (3.73) and that all functions in (3.72) and (3.73) are differentiable, with continuous second partial derivatives and nonzero Jacobian

It is essential for the theory of tensors that we allow all such changes of coordinates. In the following discussion it will be convenient to denote by ((I, . . . , 6") a jixed Cartesian coordinate system in En, in terms of which distance and angle are measured as usual. We denote by (xl, . . . , xn) and (z', . . . , 3")two other, generally curvilinear, coordiante systems introduced as before in the neighborhood D of a point, and related by Eqs. (3.72) and (3.73) to each other and related in similar fashion to the (6'). We shall refer to the (6') as standard coordinates. Now let a vector field be given in the chosen neighborhood. Then the vectors all have sets of components in each coordinate system. For a contravariant vectorjeld (or briefly, contravariant vector) the components will be denoted in the (xi) coordinates by (u' , . . . , un) (upper indices) and in the (Z') coordinates by (kl, . . . , fin). Furthermore, for any two such coordinate systems the components are to be related by the rule

This rule is chosen to fit the case when the vector in question is a velocity vector:

For then d ~ i

Of=--.-

dt

n

azi

dxj

-&x,=I

- C -uj. n

ajri

j=1

axj

For a covariant vector we denote the components in the two systems of coordinates by (u . . . , u,) and (2 . . . , fin), respectively, and require that

,,

199

200

Advanced Calculus, Fifth Edition This rule is chosen to fit the case when the vector in question is a gradient vector:

I

For then

Each contravariant vector field in the chosen neighborhood D can be obtained in the following way. One first selects a vector field U i in the ( t i coordinates-that ) . . . , ~ " ( .t. '. , defined in D . Then to each other is, n functions U 1 ( e l ., . . , t"), coordinate system, say ( x i ) , one assigns components by the equations analogous to (3.74):

,en)

Thus in ( x i ) one has similarly

Now the inverse of the matrix (axi/i3{j)is the matrix (a.$'/axj).Hence we can solve (3.76)to obtain

If we substitute in (3.77), we obtain " a x i 84, a f i at' j=1 k=l u k = xk-1E J =FI s u k

"=CCFS

n

axi

= &uk9 k=l j

so that (3.74) holds. Accordingly, once we have assigned components in the standard coordinates, we automatically obtain components in all other coordinate systems, related by (3.74), and a contravariant vector is obtained. A similar reasoning applies to covariant vector fields (Problem 5). From this discussion it follows that starting with n functions f , ( c l , . .. , defined in D, we can generate either a contravariant vector field u' or a covariant field u;. For the contravariant vector we assign components U' = f , ( e 1 ,. . . , 6") in standard coordinates and hence obtain the u' as above in all other allowed coordinates ( x i ) ;for the covariant vector we assign components U, = f,( e l , . . . , 6") in standard coordinates and hence obtain the u, in all other coordinate systems (x'). The contravariant and covariant vector fields thus obtained are related in a special way-namely, in that in standard coordinates, corresponding components are equal:

en),

Chapter 3 Vector Differential Calculus In such a case we say that the contravariant vector u' and covariant vector u, are associated. We regard the u' and u, as different aspects of one underlying geometric object u, a vector field abstracted from its components. Thus we speak of u' as contravariant components of u, u, as covariant components of u. In standard coordinates the two types of components coincide and can be identified with our usual vector components. The contravariant and covariant vectors are the tensors of Jirst order. We also introduce tensors of order zero, as scalar functions whose values are not changed by coordinate transformations. Thus in the two coordinate systems ( x ' ) and (Ti), such a function is given by f ( x l , . . . ,x n ) and ? ( T I , . . . ,i n )respectively, , where

f ( X I , .. . , x") = f ( 2 1 , . . . ,5") whenever (xi) and (f ') refer to the same point; that is,

We also call such a tensor an invariant. We can also introduce tensors of higher order. A tensor of order two requires two indices and hence can be arranged as a matrix. For example, we denote by v;, a tensor of order two that is covariant in both indices; here vi, (i = 1 , . . . , n , j = 1 , . . . , n ) are the n2 components of the tensor in the ( x i ) coordinate system. In the second coordinate system ( 2 ' ) the tensor has components Eii, and we require that

Because of the many such sums appearing in tensor analysis, one agrees to drop the sigma signs and to write (3.79) simply as

with the understanding that we sum over each index which appears more than once (k and 1 in this case). This notational rule is called the summation convention. By v f we denote a second-order tensor, called mixed, which is contravariant in the index i and covariant in the index j, and require that for a change of coordinates,

By v'j we denote a second-order tensor that is contravariant in both indices and require that

Smilar definitions are given for tensors of third, fourth, and higher orders. For example, wy denotes a tensor contravariant in i and j and covariant in k, whereby

201

f

202

4

2

Advanced Calculus, Fifth Edition The tensors of higher order appear in many geometrical and physical theories. In particular, they are needed for the basic operations on vector fields, as we shall illustrate. Each tensor of higher order can also be obtained by first assigning components in standard coordinates (6') and then using the appropriate rule to obtain components in an arbitrary coordinate system (xi). This is proved just as for contravariant and covariant vectors earlier. In particular, starting with a set of functions fiiklm(tI , . . . , t n ) , where each index i, j, . . . runs from 1 to n, one can choose some indices as contravariant and others as covariant, in some chosen order in each case, and then introduce corresponding tensor components in D, for example,

Then the rules tell us how to define the components wi? in (xi), to obtain a tensor of the type indicated. All these tensors, arising from the same set of functions in standard coordinates, are said to be associated, and we regard them as different aspects of one geometric object w. From fij(C1,. . . , e n ) we obtain in this way six associated tensors A very important tensor of second order arises in discussing distance in curvilinear coordinates. We start with standard coordinates (6') in D. For a point moving along a path in D the usual reasoning leads us to the formula

for the speed. Correspondingly, we write

(summation convention!) for the square of the "element of distance" d s . If we now introduce curvilinear coordinates (xi) but require that distance be unchanged, then we find

Hence in curvilinear coordinates, ds2 appears as a "quadratic differential form"

ds2 = gkl dxkdx',

(3.83)

where

a t i ati

=glk ( k = 1,..., n, 1 = l , ...,n). (3.84) ax' (Thus the matrix (gk[)is symmetric.) By this procedure we assign sets of components gkl in every curvilinear coordinate system. For two sets of coordinates (xi) and (xi) we have gkl=--

axk

Chapter 3 Vector Differential Calculus and hence

I.

g . .-

axk ax1 a t r a t r (a t r axk) (atr ax1) -- -----

axk ax1

-ax' a f j

a?;

a3F.i a x k ax'

axk ax1 - -a ~ an) i gk'. Therefore gij is indeed a second-order tensor, covariant in both indices. We call glj the fundamental metric tensor. We observe that the g,, reduce to (Kronecker delta) in standard coordinates, since in the (t'),

Hence we can regard the gij as the covariant tensor obtained from Sij (constant functions) in standard coordinates. From these functions we obtain six associated tensors as earlier. However, since Sji = Si,, gji = gij (as noted before), so that these two associated tensors are the same. By setting G) = S;, in standard coordinates ( t i ) we obtain a mixed second-order tensor g;. In the (x') coordinates we obtain

(since the matrices (axi/atj) and ( a t i / a x j ) are inverses of each other). Hence g!J = g!1 = Sij (constant functions)

(3.85)

in all coordinates. We often write 6) for this tensor. Next, by setting G'j = Sij in standard coordinates we obtain a contravariant second-order tensor giJ;as earlier,

Comparison with (3.84) suggests that (giJ)is the inverse matrix of (gij), that is, that

This can be directly verified (Problem 6). It follows that both matrices are nonsingular. We denote by g the determinant of (glj): g = det (gij).

(3.88)

Thus g is a scalar function, depending on the coordinate system; it is not an invariant. One can show that g must be positive (Problem 7 ) . We thus see that our fundamental metric tensor has three aspects: gij = gj; , gf = gl = S!,

g" = gii

(inverse matrix of the g;,).

(3.89)

Tensor algebra. Tensors can be combined in certain ways to yield new tensors. We can add two tensors of the same type, to obtain another of the same type (same number of contravariant and covariant indices)-for example,

203

204

4

Advanced Calculus, Fifth Edition defines the sum of the tensors vf and w:. Here we are giving the rule for a typical (xi) coordinate system and assume that v) and w) are both given in the neighborhood of a point. Thus we understand that in an (T') coordinate system,

We can now verify that the sum is indeed a tensor (Problem 8). We can multiply a tensor by an invariant; for example, from uij and f we form

and verify that this again leads to a tensor vi, of the type indicated (Roblem 9). We can multiply two tensors to form their tensor product-for example, from first-order tensors u, , v j , wk,zi we can form s;j = uivj,

tik = uiw k ,

q kl = wkz'

and can verify that these are tensors of second order of the types indicated (Problem 10). We can contract a tensor by summing over a pair of indices, one contravariant, one covariant. For example, from u i we can form wi = U:

and

v: =uG

(summed over i and j, respectively). Each such contraction lowers the order by two. We can verify that the procedure described always leads to a tensor (Problem 11). In particular, if aij, u' and vj are tensors, then

is an invariant, for ai,ukv' is a product of tensors and hence is a tensor of order four; f is obtained from this tensor by contracting twice. We can raise or lower indices of a given tensor u:: by the fonowing procedure. We multiply by the metric tensor gpq or gP4 to first increase the order by two. Then we use contraction, based on q and one selected index of u;:::, to lower the order by 2. For example, giauz is a new tensor uLij;

gka~yi is a new tensor

uyk.

We have some freedom here in where we locate the new index. However, this freedom corresponds to a choice among several associated tensors. In fact, all tensors obtained by this process are associated with the initial tensor u)i,:::.To see this, we carry out the first process in standard coordinates (6'). In these coordinates, g;, = Si,, g ' ~= Si,; u;::: becomes u;!;;; and giau$:;; becomes

But the left-hand side equals u:;.. Hence in standard coordinates we have merely moved an index to a new position. This shows that we are dealing with associated tensors.

Chapter 3 Vector Differential Calculus

Covariant and contravariant derivatives. We seek a process, analogous to forming partial derivatives, that we can apply to tensor components to yield a new tensor. In r doing so we shall add one index. We can make this covariant or contravariant and obtain corresponding covariant and contravariant derivatives. We make the process definite by simply requiring that in standard coordinates ( t i )we are in fact computing partial derivatives. s3 For example, for a (differentiable) contravariant vector u', in (xl), we obtain the corresponding U1 in then form Wf = a U' /a(J. The W; determine a mixed second-order tensor, becoming wj, in ( x ' ) , and since we have added a covariant index, we call wf, the covariant derivative of u' and denote it by AJul.It remains to calculate the w ;. We have

(c'),

Wl. J

= W ani P -a 4 a t P

p - ---a 2 t p axk axi a p u'

axj

axkax1 a t 4

atp

axj

a t p axk axi a p a d +--ax! a t 4 a p axj axk'

Here we can simplify by noting various products of inverse niatrices appearing and can write our result as

where .

The r;, are called Christofel symbols; they are not the components of a tensor. One can express them in terms of the components of the fundamental metric tensor (Problem 12):

By proceeding in this way we obtain covariant derivatives of other types of tensors. For example, we find

From these results the general pattern should be clear.

205

206

Advanced Calculus, Fifth Edition Contravariant derivatives are found in similar fashion. In fact one has the following simple rule: For every tensor A,

To justify this, we observe that the right-hand side is a tensor of the type desired and that in standard coordinates ( t i )it reduces to the set of partial derivatives of the components of A with respect to xk.Equation (3.98) expresses contravariant derivaA associated tives in terms of covariant derivatives and shows that A k A and A ~ are tensors. For an invariant f , one finds

Thus the covariant derivative of an invariant f is simply the gradient of f :

I

.d

af

Akf = grad f = axk '

3

(3.100)

The contravariant derivative of f is the associated contravariant vector. We can now define the divergence of a vector field. We want this to reduce to the usual divergence in standard coordinates. If u has contravariant components u', then we define div u = div u' = A,u",

(3.101)

since this reduces to the expected expression

in standard coordinates. Since A,uU is obtained by contraction of the tensor A ~ u ~ , it is a tensor of order 0, an invariant. We can obtain the same invariant from the covariant components of u: div u = div ui = Aau,,

(3.103)

for in standard coordinates the components become Ui = u',and again (3.102) is obtained. One can also obtain (Problem 13) the following useful expression for the divergence: 1

a

div u = --(figaAu,). 42 axh From the divergence and gradient we obtain the Laplacian of an invariant:

v2f

= div grad f = AaA, f.

(3.105)

We can also obtain this from the contravariant components of grad f :

v2f = AuAfff.

(3.106)

d

207

Chapter 3 Vector Differential Calculus Corresponding to (3.104), one has also the useful expression:

For the curl of a vector field our usual definition is specially adapted to 3-dimensional space. The appropriate generalization turns out to be a second-order tensor. We define the curl of u to be the second-order tensor

However, we verify (Problem 14) that this is equivalent to auj aui curl u = b . . - - - -. lJ - ax1 ax] Thus the form that (3.108) takes in standard coordinates is the form that it takes in arbitrary coordinates. (The same is true for the gradient operation.) By (3.109) we have

In 3-dimensional space the curl has nine components. By (3.1lo), three of these are 0, and the other six come in pairs of opposite sign:

These three values are the familiar components of the curl vector in space: au,/ay au,/az, . . . . It is because these three quantities determine the curl that the curl can be interpreted as a vector in 3-dimensional space. In 4-dimensional space, six components are needed, and one cannot associate these with a vector. For n = 2, one has essentially one component

(which is the a Q/ax - a P l a y of Green's theorem-see

Chapter 5).

Case of orthogonal curvilinear coordinates. We consider here only the case of 3-dimensional space. In Section 3.7 it is shown that for orthogonal coordinates, one has ds2 = a2dx2 p2dy2 y2dz2-that is, in the present notations,

+

+

Thus (g,,) is a diagonal matrix; its inverse is the diagonal matrix diag ( l / g l l , l/g22, l/g33), and its determinant g = gllg22g33. From (3.107) the Laplacian of f is 1

a

&af

a*'(G S )

+

1 a a a f +--1 a ---

a

ax2

( g n ax2)

aaf

-a ax3 (g33 ax3).

With gll = a 2 ,g22 = b2,g33 = y2, g = a2/?2Y2, this reduces to (3.63).

'

208

Advanced Calculus, Fifth Edition

i

!

d

One can discuss div u in the same way for the case of orthogonal coordinates. One appears to obtain a formula differing from (3.61). However, as (3.57') shows, the components p,, p,, p, of Section 3.8 are neither contravariant nor covariant components of the vector p; these components are obtained geometrically by perpendicular projection. From (3.57') we see that p u l a , p u l p , p , / y are contravariant components of p and can now verify that (3.61) is in agreement with (3.104) for the case of orthogonal coordinates. For further informationon tensors, one is referred to the books by Brand, Coburn, Hay, Levi-Civita, Rainich, and Weyl listed at the end of this chpater.

A surface S in space can be represented by parametric equations x = f (u, v), y = g(u, v), z = h(u, v). For example, x = sin u cos v,

y = sin u sin v,

z = cos u

are parametric equations of a sphere (see Problems 8-10 following Section 3.8). There are many parametrizations of a chosen surface; a second one for a sphere is X = U, y = 21, Z = (1 - u2 - v2)lf2. It is natural to allow arbitraryparametrizations,as we have done for coordinates in Section 3.9. If we are concentrating on a neighborhood D of a point on the surface, this leads us to consider all changes of coordinates from (u, v) to (u, U),(E, Z), . . . . Here for example, in a neighborhood of (uo, vo), with inverse

~

just as in Eqs. (3.72), (3.73). Indeed even though (u , v), . . . are regarded as parameters, we are here dealing with the case n = 2 of Section 3.9: that is, we have a collection of Zdimensional coordinate systems, all referring to the points of a neighborhood D of a point on the surface and related by appropriately differentiable functions. The only difference is that there are no natural standard coordinates. The absence of standard coordinates does not prevent our developing the theory of tensors. We can again do this for the n-dimensional case: that is, we consider a hypersurface of n-dimensions. This is given, for a neighborhood of a point, by a set of (parameter) coordinate systems (x', ., . , xn),(fl,. . . , .fn),. . . related by Eqs. (3.72), (3.73), as in Section 3.9. Contravariant vectors, covariant vectors and tensors in general are defined as in Section 3.9, except that one has no preferred set of standard coordinates. Typically one can measure distance on a hypersurface and hence has again ds2 = gijdxldxj,where g;, is a covariant tensor. However, in the absence of standard coordinates, gij in general never reduces to aij. We can nevertheless define g> to be Sij and g'j to be the inverse matrix of gij. We verify that gj and g'j are tensors of the types indicated. With the aid of these tensors, we can again raise and lower indices to obtain associated tensors.

Chapter 3 Vector Differential Calculus With the aid of the g;, and g'j we can dejine covariant derivatives by Eqs. (3.92), (3.94)-(3.97) and verify that these are tensors of the appropriate types. Thus it appears that, although the standard coordinates were the basis of many of the definitions in Section 3.9, we can wholly dispense with them to obtain a satisfactory theory of tensors on a surface or n-dimensional hypersurface.

*3.11 ALTERNATING TENSORS EXTERIOR PRODUCT We restrict attention to invariants and covariant tensors u ; , u i j , . . . .Each such tensor is called alternating if interchange of two subscripts changes the sign: for example, u;, is alternating if

and ujjk is alternating if for 1 5 i, j , k 5 n u1lk . . = -u

~.. , = k

-u. 1 kJ. = -U

kji

.

(3.113)

In general, the sign is reversed (unchanged) after an odd (even) number of interchanges (cf. Section 1.4 and Problem 16 following Section 1.5). Because of (3.1 12), u;, = 0 if i = j and, in general, each component having two subscripts equal is 0 . We remark that if a covariant tensor is alternating in one coordinate system, then it is alternating in each allowed coordinate system (Problem 20 following Section 3.11). Thus we are dealing with a true tensor property. For convenience we include invariants and first order covariant tensors as alternating tensors. For a tensor u;,,,, of order n , each component with n different subscripts equals fu12...,. For tensors of order m > n , all components are 0 , since of the m subscripts at least two must be equal. It is easily seen that addition of two alternating tensors of the same order produces an alternating tensor of that order, as does multiplication of such a tensor by an invariant. However, the tensor product of two alternating tensors of order at least 1 is in general not alternating (Problem 21 following). To remedy this deficiency, we observe that from each covariant tensor (order at least 2) we can obtain an alternating tensor by forming a suitable "average" of the tensors obtained from the given one by permuting indices. Thus from uij we form v.. ,, - l( u i j - u , ; ) and from uijkwe form

For a tensor ui, ;,,, of order m we form

where the sum is over all permutations of i , , . . . , i , and vj1j, is +I for an even permutation, - 1 for an odd one (cf. Problem 16 following Section 1.5); when two of i l , . . . , i , are equal, both sides of (3.1 14) are 0 . We write A ( u i j ) ,A ( u i j k ) ,. . .for the tensors formed in this way from u;, , uijk, . . . and call A the alternation operator. We also write A ( u i )= u;, A( f ) = f for orders 1

209

210

Advanced Calculus, Fifth Edition and 0. It is left as an exercise to verify that the new objects are indeed alternating tensors (Problem 22 following Section 3.1 1). With the aid of the operators A we can now obtain an alternating product of two alternating tensors. We can simply apply A to the tensor product of the two tensors. The result, slightly modified by a constant factor, is called the exterior product (or wedge product) of the two alternating tensors: If uil,,i, and vjl ,,r are alternating tensors, the exterior product is written as uil,,,;,A v,, ,,js and is defined as

Thus ui A vjk equals

This is a new alternating tensor w;jk of order 3. For more information on alternating tensors and references, see Sections 5.19 to 5.21.

PROBLEMS 1. In E~ let (el, 5') be standard coordinates and let (xl, x2) be new coordinates given by x1 = 35' + 2t2, x2 = 4 t 1 + 36'. Find the (xi) components of the following tensors, for which the components in standard coordinates are given. a) u;, where U1 = 5'e2, U2 = 6' - 5'. b) v', where V' = 6' cost2, v2 = 6' sint2. WZZ= 0. C) w;j, where Wll = 0, W12 = ( I t 2 , W2] = d) zi., where Z: = 6' + Z: = Z: = 35' + 2t2, 2: = 6' - {2. I 2. For the coordinates of Problem 1, find (a) the components of the fundamental metric tensor glj in standard coordinates and in the (xi) and (b) the gj and g'j in both sets of coordinates. 3. With reference to the coordinates and tensors of Problem 1 and with the aid of the results of Problem 2, find the following tensor components in (xi): c) wij, associated to wi,. a) ui, associated to ui; b) vi , associated to vi ;

c2,

4. a) Show that if a tensor has all components 0 at a point in one coordinate system, then all components are 0 at the point in every allowed coordinate system. b) Show that if two tensors of the same type have corresponding components equal (identically) in one coordinate system, then cooresponding components are equal in every allowed coordinate system. (One then says that the tensors are equal.) C) Show that if a tensor uij is such that u;, =-uj; in one coordinate system (xi), then iiij = Pj; in every other coordinate system (xl).(One then calls the tensor symmetric.) d) Show that if a tensor uij is such that ui, = -uj;, in one coordinate system (xi), then iiij = -ii,i in every other coordinate system (P'). (One calls such a tensor alternating, as in Section 3.11.) 5. Prove: If components U; are defined in (ti), and correspondingcomponents u; are defined in every other allowed coordinate system (xi) by the equation u; = (a6j/axi)Uj, then the components iii in (.ti)satisfy (3.73, so that a covariant vector has been defined.

Chapter 3 Vector Differential Calculus

6. Prove (3.87)from (3.84) and (3.86). 7. Prove that g = det ( g l j )is positive. [Hint: Show from (3.84) that ( g i j )is the product of two matrices and then take determinants.] 8. a ) Show that, the tensors vj, and wfi being given in a neighborhood D, Eq. (3.90)defines t ul, as components of a tensor. b ) Generalize the result of part (a) to addition of two tensors of arbitrary type. 9. Show that multiplication of all tensor components by the same invariant, as in (3.91), defines a new tensor. 10. Let u , , v,, w k ,z l , pmh be tensors. Show that each of the following tensor products is a tensor: a ) u 1 v j = si, b) u i w k = t,! C) wkzl = qkl d) wkp,h = d:,,

11. a) Let u i be a tensor of the type indicated. Show that each of the following is a tensor: w: = u ; ; / I , v ;= u ; . b ) Prove generally that contraction of a tensor yields a tensor. 12. Show from (3.93) that the rfilcan be expressed as in (3.94). [Hint: Show first that r ; [ g l S= ( a 2 ( a / a x ~ a x 1 ) ( a ( a / a x Next, s ) . with the aid of (3.84),compute the quantity in parentheses in (3.94) and show that it equals 2 r i l g l S .Now multiply both sides by gSt.l 13. Derive the formula (3.104). [Hint: First show that

+

div u = ( a u a / a x a ) r t l u l . Next put i = j = a! in (3.94), multiply out, and show that the last two terns cancel, 1 ) . obtain a g / a x l , from the determinant g, as a sum of so that rtl = ~ g a s ( a g a S / a x Now n determinants, the a t h of which is obtained from the given determinant by differentiating each element in the ath row with respect to x'. Expand each of these determinants by minors of the a t h row to obtain

gsa = A,,/g (see where A,, is the cofactor of g,, in the determinant g. Now ga" Problem 7 following Section 1.9). Thus conclude that rtl = ( 2 g ) - ' ( a g / a x l ) and show that this permits one to write div u in the form asserted.] 14. Derive (3.109) from (3.108),using (3.95). 15. We define the norm of a vector u to be its norm or length in standard coordinates. Thus for components U 1 or Ui the length is

lul = ( u ' u ' ) ~ = ( u ' u ' ) ~ . Show that Julcan be written in the forms:

lul = ( u , u ' ) f , ("1 = ( g t j u 1 u j ) 4 , lul = ( g 1 j u 1 u j ) ~ , each of which shows that the norm is an invariant. 16. The inner product of two vectors u, v is defined to be the invariant

(u, v ) = u ; v l . Show that this can be written in the forms:

21 1

212

Advanced Calculus, Fifth Edition and that (u, u) = lu12.

17. a) Prove: If u', v j are contravariant vectors and bijuivj is an invariant, then bij is a covariant tensor. b) h o v e generally: If ui , vj, .. . , wk, z', . . . are covariant and contravariant vectors and

is an invariant, then bi::: is a tensor of the type indicated.

18. a) Show that the associated tensor biJ to the curl bij of a covariant vector ui (as in Eq. (3.109)) is given by

19.

b) Show that for n = 3 in orthogonal coordinate as in Section 3.8, the three nonzero components b23,b31, b12 of the tensor b'J are the same as (3.62). [Note that the covariant components of p are a p , , pp,, yp, by (3.58').] c) Verify that bij is an alternating tensor. Show that in 3-dimensional space the exterior product w,, = u, A v, of two covariant vectors u,, v, can be interpreted in terms of the vector product. In detail, show that, in standard coordinates x , y, z with usual orientation and ul = u,, . . . , vl = v,, . . . , the . . .) x (v,i - . .) appear with or three components of the vector product (u,i signs as the nonzero components of w,, . Let u;,,,,;, be a covariant tensor in En (r 2 2). Let this tensor be altemating in one coordinate system. Show that it is alternating in each allowed coordinate system. Let u; , v; , w;, be alternating covariant tensors in E ~ . a) Show that uiv; fails to be alternating if ulvl # 0. b) Show that u; wjk fails to be alternating if ul w2l # 0. C) In one coordinate system let ui = 1 for i = 1 , 2 , 3 . Show that if u;v, is alternating, thenv; = Ofor j = l , 2 , 3 . Show that if u;, ;, is a covariant tensor, then Eq. (3.114) defines a new covariant tensor v;, . im which is altemating.

+

20. 21.

22.

+

+

Differential geometry of surfaces in space. In the following problem let S be a surface in space as in Problems 8-10 following Section 3.8. Thus one has curvilinear coordinates (u, v) on S and a positive definite quadratic differential form ds2 = E du2 2 F du dv G dv2, called the frstfundamental form of S. If we write u1 = u, u2 = v, gll = E , g12 = g21 = F , g22 = G. then ds2 = g;, dui duJ and, as in Section 3.10, gij is a covariant second order tensor. Also g = det (g;;) = E G - F~ > 0. As in Problem 16 following Section 2.13 a unit normal vector for S can be obtained as n = w/ 1 wl, where w = ar/au x ar/a v. The choice of normal n (varying from point to point) will be assumedfxed and unaffected by changes of coordinate on S. (We are here dealing with an oriented surface as in Section 5.9.)

+

+

23. Let a path C on S be given by parametric equations u = u(t), v = v(t) in curvilinear coordinates, so that r = r(u(t), v(t)) is the vector equation of the path. We assume that t is arc length s on C and that appropriate derivatives exist, so that the Frenet formulas

Chapter 3

213

Vector Differential Calculus

of Problem 6 following Section 2.13 apply. We write r,, r,, r,,, r,,, r,, for partial derivatives, all assumed to be continuous. a) Show that the unit tangent vector T = d r l d s satisfies the equation T = (du/ds)r, (dv/ds)r,, where

+

b) Show that the acceleration vector a = d T l d s satisfies the equation

c) Show that the curvature K satisfies the equation

where L = r,, . n, M = r,, - n, N = r,, . n. The quadratic differential form L du2 2M d u d v N dv2 is called the second fundamental form of S. d) Write L l l = L , L I 2 = L21 = M ,L22 = N and show that Lij is a tensor as indicated. [Hint: Fix a point Po: (uo,vo) on S and choose a special basis i l ,j l , k l for vectors in space with k l = n at Po. Then write r = ro til r]jl J k l , with (6, r ] , J ) = (0,0,O)at Po. Show that at Po r, .n = 0 and r , .n = 0 , so that at (uo,vo) J, = 0 and 5; = 0 and L = J,,, M = <, N = 5;, These conditions must hold at Po for each allowable coordinate system ( u , v). For new coordinates ( f i ,5 ) with ii = fio, O = 00 at Po, show by the rules of Section 2.16 (Eq. (2.133) and Problem 6 following that section) that at Po

+

+

+ + +

t u = Jfifi(fiu)2

+ 2Jfiofiu9,+ Joo(Ou)

and obtain analogous formulas for J,, and J,,. become

2

Show that in tensor notation these

24. Normal curvature on S. The results of Problem 19(c) show that all paths C on S through a chosen point Po with fixed tangent vector T and principal normal N at Po, have the same curvature K at Po. Hence it suffices to consider plane sections of S. We consider such sections by planes through Po, other than the tangent plane to S at Po, let 0 = 6 ( N , n) for each path C obtained as such a plane section. a ) Show that for fixed T and variable N one has K cos 8 = const. at Po. Here the constant has absolute value K,, where K,, called the normal curvature of S , is the curvature at Po of the section of the surface containing the tangent line to the path at Po and the normal line to the surface at Po. [Thus K = fK,; the f sign follows from the fact that either 0 5 8 < n / 2 or n / 2 < 0 5 n for all N.] b) Show that

25. Mean and Gaussian curvatures. Seek the maximum and minimum of the normal curvature K, for variable tangent vector T at a fixed point Po. [Hint: Let = d u l d s , q = dvlds. Then by Problem 24(b) K, = I L <+~2 M t q ~ q and ~ one 1 is led to seek the

+

<

*

214

5

Advanced Calculus, Fifth Edition

+

.{

+

+

+

critical points of f (6, q) = L S ~ 2M6q ~ q ' ,where Ec2 2Feq Gq2 = 1. Use Lagrange multipliers as in Problems 14, 15 following Section 2.21 to obtain the quadratic equation L-IE M-IF

M-AF N-IG

with real roots I15 1 2 and corresponding eigenvectors vl = (61, ql), v2 = (e2, q2). Conclude that f has an absolute minimum IIand absolute maximum A2 under the side condition. One terms H = ;(I1 h2) the mean curvature and K = A l A 2 the Gaussian curvature of the surface at Po. By Problem 14(b) following Section 2.21, if I1# h2, then

+

I

and hence the corresponding tangent vectors are orthogonal (Problem 10(e) following Section 3.8). The corresponding directions, or their opposites, are called directions of principal curvature. If hl = hz, f (6, q) has a constant value Il on the ellipse E t 2 2F5q Gq2 = 1 and hence f = Al(Et2 2FCq Gq2) for all (6, q); if further Al # 0, then the point Po is called an umbilical point of S-here all plane sections of S through Po have the same curvature IAl I; if I1= A2 = 0, then Po is called aparabolic umbilic of S.]

+

I

+

+

+

Suggested References Brand, Louis, Vector and Tensor Analysis. New York: John Wiley and Sons, Inc., 1947. Coburn, N., Vector and Tensor Analysis. New York: Macmillan, 1955. Gibbs, J. Willard, Vector Analysis. New Haven: Yale University Press, 1913. Hay, G. E., Vector and Tensor Analysis. New York: Dover, 1954. Levi-Civita, T.,The Absolute Differential Calculus, transl. by M. Long. London: Blackie and Son, 1927. Phillips, H. B., Vector Analysis. New York: John Wiley and Sons, Inc., 1933. Rainich, G. Y., Mathematics of Relativity. New York: John Wiley and Sons, Inc., 1950. Especially Chapters 1,2, and 4. Struik, Dirk J., Lectures on Classical Differential Geometry, 2nd ed., Reading, Mass.: Addison-Wesley, 1961. Weyl, Hermann, Space, Time, Matter, transl. by H. L. Brose. New York: Dover, 1952.

I

Integral Calculus of Functions of Several Variables

We review here the properties of the integral for functions of one variable. Let f ( x ) be defined for a 5 x 5 b. The definite integral

is defined as a limit:

Here one is considering subdivisions of the interval a 5 x 5 b by values a = xo < < ~ 2- - .< X , = b, and A i x = xi - x i - ] , xi-] 5 xf 5 x i , h is the largest of Alx, . . . , A , x ; we call h the mesh of the subdivision. The limit (4.1) is said to exist and have value c if for every E > 0, one can choose 6 > 0 so small that for every such subdivision of mesh h less than 6 and no matter how the x: are chosen in the intervals x i - , 5 x 5 x i , one has

XI

The integral we have defined is called the Riemann integral. It can be shown to exist if, for example, f ( x ) is continuous for a 5 x 5 b; see Section 4.10. Further, it exists if f is continuous except at a finite number of points, provided that f is bounded: If (x)l 5 M for some constant M ,for a 5 x 5 b. 215

216

Advanced Calculus, Fifth Edition From the definition we deduce some basic properties of the integral (see CLA, Chapters 4 and 13):

Ja

ifA If ( x ) B, ~ a j x (b.

(4.7)

These are all valid if, for example, f is continuous for a 5 x 5 b. We call F(x) aprimitive or antiderivative off (x) or an indejnite integral off (x) C, if F1(x) = f (x) on the interval. All such primitives are expressible as F(x) where F(x) is one primitive and C is an arbitrary constant. One writes

+

to describe the relationship. If f is continuous on the interval, then one primitive o f f is given by

Thus one has the rule, often called the fundamental theorem of calculus,

The function F(x) can be interpreted as the area under the graph of f from a to x, as in Fig. 4.1, provided that f is positive.

Figure 4.1

Function F ( x ) of Eq. (4.9) as area.

. Chapter 4

217

Integral Calculus of Functions of Several Variables

If a primitive F of f is known for a 5 x 5 b , then

Lb

l

b

f ( x )d x =

F 1 ( x )d x = F ( x ) ~ = : F(b) - F(a).

(4.11)

This rule is often considered to be part of the fundamental theorem of calculus. It is the basis for the familiar procedures of elementary calculus; for example, cosn 1 +-=I. 2 2 The average value or mean value of f on the interval a 5 x 5 b is 1 b-a

If f ( x ) is continuous on the interval, then for some x* on the interval, f ( x * ) equals the mean value; that is,

This is the Mean Value theorem for integrals. If a continuous function f is such that

lfl

f(x)dx =0

(18

(4.14)

for every interval a1 5 x g bl contained in the given interval a 5 x 5 b , then f ( x ) I 0. This follows from the Mean Value theorem (Problem 11 following). As a consequence of this result, we deduce that if f is continuous and f (x) 2 0 for a 4 x 5 b. then

f(x)dx =0 implies that f ( x ) = 0 (see Problem 12 following). For b < a , one defines:

and for a = b the integral is defined to be 0.One then verifies that for f continuous on an interval containing a , b, c,

The rule (uv)' = uv' + u'v leads to a procedure for integrating by parts, which we state for definite integrals:

, are continuous for a 5 x 5 b. Here we assume that u ( x ) ,v ( x ) ,u f ( x ) v'(x)

218

Advanced Calculus, Fifth Edition

3 If the Riemann integral of f fails to exist, one may still be able to obtain a value by treating the integral as an improper integral. For example,

does not exist as a Riemann integral for p > 0, since the integrand is unbounded, and the integral is improper because of the discontinuity at x = 0. One seeks a value by a limit process

For p = 1 the integral from b to 1 has the value -log b, so that there is no limit as b + 0. Otherwise, the integral has value

@

1-P and this has limit 1 /(I- p ) for 0 < p < 1 but has no limit for p > 1. Hence one writes

and assigns no value for p 2 1 . Here the improper integral is said to converge for 0 < p < 1 and to diverge for p > 1 . If there are several discontinuities (a finite number) on the interval, one breaks the integral up into integrals with just one discontinuity; if each of these converges, then the given integral is termed convergent and assigned as value the sum of the values of the partial integrals; otherwise, it is termed divergent. For example, for an integral from 0 to 5 with discontinuities at 1 and 4, one considers the integrals from 0 to 1, 1 to 3, 3 to 4 , and 4 to 5. Integrals with infinite limits are also improper, and one evaluates them by a limit process. For example, 1 w

Lm

0

xe-" d x =

K

J__ xe-x2 dx + e - ~ 2 0

= lim b-,-w

-2

.

dx = lim arc tan b = -, b-t m 2

l

03

xc"

dx *

+ b+m lim -2

:t

I

.

'1

Improper integrals are discussed further in sections 6.22 to 6.24. Numerical evaluation of dejnite integrals. The formula (4.11)is often of no help, since the primitive F is hard to find (or may not exist in usable form). Hence in practice, many definite integrals are evaluated numerically with the aid of computers. Sophisticated procedures to this end have been developed. We mention here only

4

219

Chapter 4 Integral Calculus of Functions of Several Vaiiables

Trapezoidal rule for

Figure 4.2

sobf ( x )d x .

one simple rule, which is often sufficiently accurate, the trapezoidal rule:

Here one has a subdivision of the interval as above. The right-hand side can be interpreted as replacing the integrals from xi-, to xi by areas of trapezoids, as in Fig. 4.2. When all A i x are equal, the rule becomes

lbf(x)dx

5

b-a - [2n f(a)

+ 2 f ( x 1 )+ 2f(x2)+ - - +. 2 f ( x n - d + f(b)].

(4.20)

If I fr'(x)l 5 L = const on the interval, then one can show that the sum on the right-hand side of (4.20) differs from the integral by at most L(b - ~ ) ~ / ( 1 2 (See n~). CLA, pp. 561-562.)

PROBLEMS 1. Evaluate the following indefinite integrals:

a) j x 2 s i n x d x c)

b)

dx S(X-,)(x-2)1

1< X ( 2

d,

I* x z l

I+-* d X

1

2. Evaluate the following definite integrals:

b)

a) jd d C 7 d x

c) jd(2x2 - 3x

+ l)eX dx

1 : sin 2x sin 3x d x

d) J: arctanx d x

3. Show that the following integrals converge by evaluating the corresponding limits: 1 dx b)b ,jOCxdx , c ) jd log x d x , a) Jo m, d)

J?

dx

e)

- 9

4. Evaluate where possible:

a)

1

dx

b)

7J -

d) ; 1 :; : 3 $

dx,

e)

jr

x2e-'

dx,

5. ;j sin x d x ,

0 Jp" c)

17

y 3

dx

9

f ) jooO(l- tanhx)dx

220

Advanced Calculus, Fifth Edition

a*'

',

I

5,.

/

5. Find the area between the curves: a) y = O and y = 1 - x 2 , = x 3 and = x t , C) y = 6arc sinx and y = n s i n nx. 6. Find the average value of the function given over the interval indicated: a) f ( x ) = s i n x , O f x i ; b)

b) f ( x ) = sinx, C)

-5

.B

~x 5 0 ;

R f(x) = sin2x, 0 5 x 5 T ;

+

d) f (x) = a x b, X I i x f x2. 7. Prove: Iff (x) and g(x) are continuous for a 5 x Ib and (g(x)- f (x)l I r fora then Ilbg(x)dx -

x Ib,

lb

f ( r ) d x ( f r(b - a).

[Hint: Use (4.6).] 8. On the basis of Problem 7, use the suggested approximation to evaluate each integral and estimate the error: sin (x2)dx, approximate sin (x2) by x 2 - (x6/6); a)

[d b) [d

1 t

+ $.

e-" dx, approximate r-"' by 1 - x 2 9. Show that if f ( x ) is continuous for 0 5 x 5 1, then

10. Using the result of Problem 9, show that

).,

lim

lP+2P+..,+nP

n+w

d) n+m lim A{(n

,, P+I

- 1 - P+1' P L O

+ l)(n + 2) .(2n)) A = $ [Hint: Use f (x) = log (1 + x).]

11. Prove: I f f (x) is continuous for a

f

%1

x 5 b and

I:'

f (x) dx = 0

for every interval a1 5 x 5 bl contained in the interval a 5 x 5 b, then f (x) = 0. [Hint: Fix xo and apply the Mean Value theorem (4.13) to the interval xo < x 5 xo 6 (or xo - 6 f x 5 x0) for small positive 6. Now let 6 + 0.1 12. Prove: If f (x) is continuous for a p x 5 b, f (x) 2 0 on the interval, and

+

Ib f (x) d x = 0.

Ja

then f (x) = 0. [Hint: Take c, a < c < b, and observe that the integrals from a to c and from c to b must be positive or 0, but their sum is 0, so both must be 0. Conclude that ja:' f (x)dx = 0 for every choice of a,, bl on the interval a 5 x 5 b and hence, by Problem 11, f (x) = 0.1

Chapter 4 Integral Calculus of Functions of Several Variables

It is shown in advanced mathematics that the indefinite integral

cannot be expressed in terms of elementary functions. It is, however, possible to evaluate the definite integral

as accurately as desired, by the methods mentioned in Section 4.1. We shall now see that a similar statement holds for the indefinite integral. The procedure to be given is an application of the fundamental theorem

of Eq. (4.10). As was pointed out in Section 4.1, the theorem implies that the equation

defines a function F(x) such that F1(x) = f (x); that is, F(x) is an indefinite integral of f (x); the complete indefinite integral is then given by

1

f (x) dx = F(x)

+C.

The function F(x) can be numerically evaluated, for eachfiedx, by the procedures mentioned at the end of the preceding section. If this is done for a sequence of values of x covering the range a 5 x ( b, F(x) will be known in tabular form and can be graphed. It is in this sense that the indefinite integral is being numerically evaluated.

EXAMPLE To evaluate

for 0 5 x 5 1, one computes

for x = 0,0.1,0.2, . .. , 1. The work is shown in Table 4.1. The integrals are computed by the trapezoidal rule. The fourth column, which shows F(x), is the "cumulative sum" of the third column. Thus the third column gives the differences A F,

221

222

Advanced Calculus, Fifth Edition TABLE 4.1 x

esLn

+O;]

' es'"'dt

1 ; eslnrdt= F ( x )

and if these are divided by A x = 0.1, one should obtain f ( x ) again approximately. For example,

This gives f ( x ) for x about equal to 0.55. The complete indefinite integral (4.22) is now given by F ( x ) C, where F ( x ) is given by Table 4.1, for 0 5 x ( 1. The range of x can clearly be extended as far as desired and the accuracy improved as required, so that the method should meet all practical needs.

+

Elliptic integrals. An integral of the form

is an example of an elliptic integral. It can be shown that this integral cannot be evaluated in terms of elementary functions. Accordingly, some numerical method is called for. Because of the importance of this integral in applications, elaborate tables have been computed of the function

for various values of the constant k. Such tables are given in A Short Table of Integrals, 3rd ed., by B. 0.Peirce (Boston: Ginn, 1956)and Tables of Higher Functions 6th ed. revised by F. Losch, E. Jahnke and F. Emde (New York: McGraw-Hill, 1960). .>I.

7

Chapter 4 Integral Calculus of Functions of Several Variables The integral (4.24) is called an elliptic integral of first kind. Integrals of the second and third kinds are illustrated by

where 0 < k2 < 1 , a # 0, a2 # k 2 . The integral (4.25) arises in computing the length of an arc of an ellipse (Problem 4), and this is the basis of the term elliptic integral. It can be shown that if R ( x , y) is a rational function of x and y (Section 2.4) and g(x) is a polynomial in x of degree 3 or 4, then the integral

can be expressed as an elementary function plus elliptic integrals of the first, second, or third kind. Accordingly, numerical evaluation of a few integrals permits precise evaluation of a broad class of integrals. This puts additional emphasis on the fact that every indefinite integral determines a function of x. For each new indefinite integral that we study and evaluate, a broad class of new functions becomes both numerically useful and open to a thorough analysis. Indeed, the trigonometric functions could have been dejned in such a manner:

defines y = arc sinx and hence x = sin y. This is more awkward than the usual geometric definition, but all properties can be deduced from such a definition (cf. Problem 3). For further information on elliptic integrals we refer to the books by von K h i i n and Biot (Chapter 4) and Whittaker and Watson (Chapters 20-22) listed at the end of the chapter. Some properties are obtained in Problems 5 to 7.

PROBLEMS

b.

1. Evaluate numerically: a) J x d x f o r O ~ x <10,usingx=0,1,2,.... b) ~ e - x 2 d x f o r ~ _5( 1,usingx x =0,0.1,0.2 ,.... C)

Jcosxdx, O s x I 1,usingx =0,0.1,0.2 ,.... O s x 5 1,usingx=O,O.1,0.2,....

d) J - & d x , l+x

e) J m d x , O ~ x ~ 0 . 5 , u s i n g x = 0 , 0 . 1 , 0,.... .2 The answers should be graphed and checked by differentiation.

;=

zf

224

,:nl

Advanced Calculus, Fifth Edition

2. Let f ( x )be continuous for a a) Prove that

jx

r i

.

*

-

t

t

,

. .*-. r'

Lb

.bf

5 b.

t

t =x

a

jx

5 b.

b ) Prove that

2 lx2

1

f ( t ) d t = 2 x f ( x 2 ) , a 5 x2 j b.

d C)

S"'Hmt: ! Let u = x2,

SO that the integral becomes a function F(u). Then, by the chain rule, ad;;F(u) = F'(u)&. ] Prove that

B

j'

d) Prove that

,

[Hint: Let u = x 2 , v = x3. Then the integral is F(u, v ) . By the chain rule,

3. The function log x (base e always understood) can be dejined by the equation 10gx =

ix

, x>o.

(a)

a ) Use this equation to evaluate log 1, log 2, and log 0.5 approximately. b ) Prove, from Eq. (a), that log x is defined and continuous for 0 < x < oo and that

d-logx - -1 dx x' C) Prove that log (ax) = log a +log x for a > 0 and x > 0. [Hint: Let F ( x ) = log (ax)log x. Use the result of (b) to show that F1(x)= 0, so that F ( x ) = const. Take x = 1 to evaluate the constant.] 4. Let an ellipse be given by parametric equations: x = a cos 4, y = b sin 4, b > a > 0. Show that the length of arc from 4 = 0 to 4 = cw is given by

1

5. Show that the function F ( x )defined by (4.24) has the following properties: a) F ( x ) is defined and continuous for all x ; b ) as x increases, F ( x ) increases; c) F(x n) - F ( x ) = 2K, where K is a constant; d) lim F ( x ) = oo, lim F ( x ) = -oo.

+

X+W

X-+-ca

5

;

Chapter 4 Integral Calculus of Functions of Several Variables 6. The function x = a m ( y ) is defined as the inverse of the function y = F(x) of (4.24). Uslng the results of Problem 5, show that a m ( y ) is defined and continuous for all y and has the following properties: a) as y increases, a m ( y ) increases: b ) am(y 2 K ) = am(y) n;

+

c)

3= Ji-zGG.

+

7. The functions s n ( y ) , c n ( y ) , d n ( y ) are defined in terms of the function of Problem 6 by

the equations: s n ( y ) = sin [ a m ( y ) ] ,

c n ( y ) = cos [ a m ( y ) ] ,

dn(y)=

Prove the following identities: a) s n 2 ( y )+ c n 2 ( y )= 1;

d

m j_

I

. r

b) $ s n ( y ) = c n ( y ) d n ( y ) : C)

d

a ; ~ n (= ~ )- s n ( y ) d n ( y ) ;

+ + +

4

1

d) sn(y 4 K ) = s n ( y ) ; el cn(y 4 K ) = c n ( y ) ; f) dn(y 2 K ) = dn(y). The functions s n ( y ) , c n ( y ) , d n ( y ) are called ellipticfunctions. It should be emphasized that they depend on k, in addition to y . 8. The error function y = erf (x) is defined by the equation

This function is of great importance in probability and statistics and is tabulated in the books mentioned after (4.24).Establish the following properties: a) erf ( x ) is defined and continuous for all x ; b ) erf ( - x ) = -erf ( x ) ; c ) -1 < e r f ( x ) < 1 forallx.

lab

The definite integral f ( x ) d x is defined with respect to a function f (x) defined over an interval a 5 x 5 b. The double integral

will be defined with reference to a function f ( x , y ) defined over a closed region R of the xy-plane. It is necessary further to assume that R is bounded, that is, that R can be enclosed in a circle of sufficiently large radius; otherwise, just as in the case when a or b is infinite, the integral is improper. The definition of the double integral parallels that of the definite integral. One subdivides the region R by drawing parallels to the x and y axes, as in Fig. 4.3. One considers only those rectangles that are within R and numbers them from 1 to n, denoting by A, A the area of the ith rectangle and by h the maximum diagonal of

225

226

Advanced Calculus, Fifth Edition Y

,

Figure 4.3

The double integral.

Figure 4.4

Reduction of double integral to iterated integral.

the rectangles; h is called the mesh. One then forms the sum

where ( x f , yf) is chosen arbitrarily in the ith rectangle. If this sum approaches a unique limit as the maximum diagonal h approaches 0, then the double integral is defined as that limit:

The existence of the double integral can be demonstrated when f is continuous and R satisfies simple conditions, in particular when R can be split into a finite number of pieces, each of which is described by inequalities of the form or of the form where the functions y l ( x ) , y 2 ( x ) , x ~ ( y )x, ~ ( y are ) continuous. Figure 4.4 illustrates the first form.

Chapter 4 Integral Calculus of Functions of Several Variables

Figure 4.5

The double integral as volume.

pl

Just as the definite integral of a function f (x) can be interpreted in terms of area, so can the double integral be interpreted in terms of volume, as suggested in Fig. 4.5. Here f (x, y) is assumed to be positive over the region R (the same type as in Fig. 4.4), and the volume being computed is that beneath the surface z = f (x, y) and above the region R in the xy-plane. Each term f (x:, y:) A, A in the sum (4.30) represents the volume of a rectangular parallelepiped with base A i A and altitude f (x:, y;), as shown in Fig. 4.5. It is clear that the sum of these volumes can be regarded as an approximation to the volume beneath the surface and that the double integral (4.30) can be used as definition of the volume. The evaluation of double integrals for a region like that of Fig. 4.4 is facilitated by reduction to an iterated integral:

For each fixed value of x the inner integral

is simply a definite integral with respect to y of the continuous function f (x, y). This integral can be interpreted as the area of a cross section, perpendicular to the x axis, of the volume being computed; this is suggested in Fig. 4.5. If the area of the cross section is denoted by A(x), then the iterated integral gives

If this integral is interpreted as a limit of a sum of terms C A(x) Ax, then it is geometrically evident that the integral does represent the volume.

227

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Advanced Calculus, Fifth Edition

THEOREM If f ( x , y ) is continuous in a closed region R described by the inequalities (4.31), then, for x l i x 5 x2,

is a continuous function of x , and

;d

Similarly, if R is described by (4.32),

4

For a proof of existence of the double integral of a continuous function and of the reduction to an iterated integral, the reader is referred to Section 4.11. EXAMPLE Let R be the quarter-circle, 0 5 y 5 f ( x , y ) = x 2 y2. Then one has

+

1/-,

0 5 x 5 1 , and let

The double integral can be interpreted and applied in a variety of ways. The following are illustrations:

I. Volume. If z = f ( x , y ) is the equation of a surface, then

gives the volume between the surface and the xy-plane, volumes above the xy-plane being counted positively and those below thexy-plane being counted negatively. 11. Area. If one takes f (x, y ) E 1, one obtains

Chapter 4 Integral Calculus of Functions of Several Variables

111. Mass. If f is interpreted as density, that is, as mass per unit area, then M = mass of R =

(4.37) R

IV. Center ofmass. If f is density, then the center of mass ( 2 , J) of the thin plate represented by R is located by the equations

where M is given by (4.37). V. Moment of inertia. The moments of inertia of the thin plate about the x - and y-axes are given by the equations

while the polar moment of inertia about the origin 0 is given by

The basic properties of the double integral are essentially the same as those for the definite integral:

where R is composed of two pieces R1 and R2 overlapping only at boundary points;

where A is the area of R, as in (4.36), and (x,, y , ) is a suitably chosen point of R ;

229

230

Advanced Calculus, Fifth Edition if MI 5 f ( x , y ) IM2 for ( x , y ) in R , then

,

F

where A is the area of R . The functions f ( x , y) and g(x, y ) are here assumed to be continuous in R. The proofs of these properties are essentially the same as for functions of one variable and will not be reviewed here. From (4.45)one can prove the inequality: //f(x,y)dxdy

cM

. 4

1

(4.47)

R

where If ( x , y)I 5 M in R. Equation (4.44) is the Mean Value theorem for double integrals. If both sides are divided by A, one obtains

The right-hand side can properly be termed the average value o f f in R [cf. Eq. (4.12)]. The Mean Value theorem thus asserts that a function continuous in R takes on its average value somewhere in R . If R is a circular region R, of radius r with center at (xo,yo),then one can consider the effect of letting r approach 0 in (4.48).The region R, is thus being shrunk down to the point (xo,y o ) Since ( x , ,y , ) is always in R,, it must approach (xo,yo);since f is assumed to be continuous, it must approach f (xo,yo). Accordingly,

where A, = n r 2 = area of R. The operation on the left-hand side suggests what might be called "differentiation of the double integral with respect to area." Thus the derivative, in this sense, of the integral reproduces the function being integrated. Accordingly, (4.49)parallels the theorem

for functions of one variable. It should be remarked that the "incremental area" R, of (4.49)could also be chosen as a rectangle, ellipse, and so on. Equation (4.49)has a physical interpretation, namely, the determination of density from mass. Thus if a thin plate of variable density is given, one can measure the mass of the plate or any portion of it directly, simply by weighing the portion concerned. To determine the density at a particular point of the plate, one chooses a small area about this point, measures its mass, and divides by its area. One is thus carrying out experimentally a stage of the limit process (4.49)in the form mass lim -= density. area

9

Chapter 4 Integral Calculus of Functions of Several Variables The following property of double integrals is established as for single integrals (Problems 11 and 12 following Section 4.1 ; see also Problem 14 following Section 2.18): Let f (x, y) be continuous in the bounded closed region R. If

for every bounded closed region RI contained in R, then f ( x , y) f ( x , y) L 0 in R and

then f (x, y)

E

0 in R. If

= 0 in R.

Remarks on the definition of the double integral. For a subdivision such as that of Fig. 4.3, one can include in the sum (4.29) a term for each rectangle partly included in R , using a point ( x f , y:) of R in the rectangle and taking as A, A the area of the irregular piece. For the type of region considered here, the total contribution of such terms can be shown to be negligible in the sense that for h sufficiently small, their sum has an absolute value that is less than a prescribed positive 6 . The same assertion holds if one includes the boundary pieces and in each case uses as Ai A the area of thc whole rectangle. One can also subdivide R by curves other than straight lines, as in Fig. 4.6. Under some natural restrictions on these curves and on R the areas A I A, A2A, . . . are well determined, and the sum (4.29) can be formed. The mesh h is now the maximum "diameter" of the n pieces (see Fig. 4.6). If f is continuous in R , then Eq. (4.30) remains valid. Because of the freedom to use arbitrary shapes, one also writes the double integral as

Figure 4.6

General subdivision for a double integral.

231

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Advanced Calculus, Fifth Edition

4.4

TRIPLEINTEGRALS

, . +

3

AND MULTIPLE INTEGRALS IN

GENERAL

The notion of double integral generalizes to integrals of functions of three, four, or more variables:

These are called tGple, quadruple, . . . integrals. In general, one terms them multiple integrals, the double integral being the simplest multiple integral. For the triple integral, for example, one considers a function f ( x , y , z ) defined in a bounded closed region R of space. One subdivides R into rectangular parallelepipeds by planes parallel to the coordinate planes, numbers the parallelepipeds inside R from 1 to n, and denotes the ith volume by Ai V. The triple integral is then obtained as the limit of a sum:

a

here h is the mesh, the maximum diagonal of the A i V . The point ( x f , y f , zf) is arbitrarily chosen in the ith parallelepiped. The existence of a unique limit can be shown if f ( x , y, z ) is continuous in R. The simplest theory holds for a region R described by inequalities such as the following: For this region, one can reduce the triple integral to an iterated integral by the equation

as in the theorem of Section 4.3. EXAMPLE

If R is described by the inequalities 05x51,

05y5x2,

3

o ~ z ~ X + y

and f = 2 x - y - z, one has

The limits of integration can be determined by the following procedure. Let the order chosen be d z dy d x as in the preceding example. Then we determine the

Chapter 4 Integral Calculus of Functiodi of Several Variables x limits x l , x2 first as the smallest and largest values of x on the figure R. We then consider a cross section: x = const of the figure, taken between x l and x2. The y limits are now determined as the minimum and maximum values of y in the cross section; these depend on the x chosen and are hence yl(x) and y2(x). Finally, within the cross section we find the minimum and maximum values of z for each fixed y; these are the z limits; they depend on the constant values of x and y chosen and are hence zl(x, y) and z2(x, y). Although the minima and maxima referred to can often be read off a sketch of R as a 3-dimensional figure, it is usually easier to plot only typical cross sections as 2-dimensional figures. Since the definite integral can be interpreted as area and the double integral as volume, one would expect the triple integral to be interpreted as "hypervolume:' or volume in a Cdimensional space. Although such an interpretation has some value, it is simpler to think of mass; for example,

JSS

f (x, y, z) dx dy dz = mass of solid of density f.

The other interpretations of the double integral also generalize:

//I

dx dy dz = volume of R;

where (T, 7, Z) is the center of mass;

t 3ns t

R

the moment of inertia about Ox. The basic properties (4.41) to (4.47) also generalize to all multiple integrals. Thus, in particular, one has the Mean Value theorem:

where V is the volume of R. This can also be made the basis for "differentiating a triple integral with respect to volume," as in (4.49).

Let

F(t) = f (t)i

+ g(t)j + h(t)k,

a 5 t 5 b,

be a continuous vector function of t, so that f (t), g(t), h(r) are all continuous for

233

1

234

a 5r

9

, '

Advanced Calculus, Fifth Edition

:I

* ‘

.s,l,

5 b. We then define:

lb

F(t)d t =

lb

f( r ) dt i

+

lb

g(r) dr j

+

Ih

h(r)dr k.

(4.57)

a

This integral can be interpreted as the limit of a sum:

as in (4.1).The limit is a vector c and existence of the limit means that given r > 0, there is a 6 > 0 such that, for 0 < h < 6 ,

The properties (4.3), . . . , (4.6),(4.10),(4.16)all extend at once to the integral of F(t),where the absolute value If (x)l is replaced by the vector norm (F(t)l. One can also integrate vector functions of several variables. For example, if

is continuous on the bounded closed region R, suitable for triple integrals, then

This integral can also be interpreted as the limit of a sum, and the familiar properties of triple integrals carry over.

1

PROBLEMS 1. Evaluate the following integrals: a) j j R ( x 2+ y 2 ) d xd y , where R is the triangle with vertices (0, O), (1, O), (1, 1);

b) j j j R u2v2wdu d v d w ,where R is the region: u2 C)

3t

+ v 2 5 1,0 5 w 5 1 ;

S f Rr3 cos 0 dr do,where R is the region: 1 5 r 5 2, 2 5 0 5 n ;

d) f f fR(x + 2 ) d V , where R is the tetrahedron with vertices (O,O, O), ( 1 , 0 , 0 ) ,( 0 , 2 ,O), ( O , O , 3). 2. Find the volume of the solid region below the given surface z = f ( x , y ) for ( x , y) in the region R defined by the given inequalities: a) z = e X c o s y , 0 1 x 5 1 , O 5 y 5 n / 2 b) Z = X ~ ~ - ~ - YO, ~ < X1 , 0 5 y 5 2

f #

Chapter 4 Integral Calculus of Functions of Several Variables

c) z = x Z y , 0 5 x 5 1 , x + l i y i x + 2 d) z = , / m - , x2-y2,~, O(x< 1 3. For each of the following choice of R, represent JJ f ( x , y ) d A over R as an iterated integral of both forms (4.33) and (4.34): a) 1 5 x 5 2 , 1 - x i y l l + x b) y2+x(x- 1 ) ( 0 4. Evaluate j j j R f ( x , y, z ) d V for the following choices of f and R: a) f ( x , y, z ) = J-, R the cube of vertices (O,0, O), ( L O , O),(0,1, O), (0,0, I), (1, 1,O). ( l , O , I), (0, 1, l ) , (1. 1, 1). b ) f ( x , y, z ) = x2 + z2, R the pyramid of vertices ( f 1, f 1 , O ) and ( 0 , 0 , 1). 5. For each of the following iterated integrals, find the region R and write the integral in the other form (interchanging the order of integration): a)

I!

j,',2&-X f ( x , y ) d y d x

1

I+*

d ) So j,-, f ( x , y ) d y d x 6. Express the following in terms of multiple integrals and reduce to iterated integrals, but do not evaluate: a) the mass of a sphere whose density is proportional to the distance from one diametral plane; b) the coordinates of the center of mass of the sphere of part (a); C ) the moment of inertia about the x-axis of the solid filling the region 0 5 z 5 1 x2 - y 2 , 0 5 x 5 1 , O 5 y 5 1 - x and having density proportional to xy. 7. The moment of inertia of a solid about an arbitrary line L is defined as

where f is density and d is the distance from a general point ( x , y, z ) of the solid to the line L. Prove the Parallel Axis theorem: where is a line parallel to L through the center of mass, M is the mass, and h is the distance between L and Z . , ( ~ i n t Take : to be the z axis.) 8. Let L be a line through the origin 0 with direction cosines 1, m, n. Prove that

where Ixy =

/fl

x y f ( x , y, z ) d x dy dz.

R

-

I,, =

/fl

yzf

-

R

The new integrals are called products of inertia. The locus

Ixx2

+ l y y 2+ zzz2 - 2(Ixyxy + I,,yz + Izxzx) = 1

is an ellipsoid called the ellipsoid of inertia.

. .. .

J

236

Advanced Calculus, Fifth Edition

I*

r;r!

I

9. It is shown in geometry that the medians of a triangle meet at a point, which is the centroid of the triangle, and that the lines from the vertices of a tetrahedron to the centroids of the opposite faces meet at a point which is 314 of the way from each vertex to the opposite face along the lines described. Show that this last point is the centroid of the tetrahedron. [Hint: Take the base of the tetrahedron to be in the xy-plane and show that ? = h/4, if h is the z-coordinate of the vertex not in the xy-plane.] 10. Evaluate the integrals:

a)

& F ( t ) d t ,if F(t) = t2i,- e'j + &k.

b)

SIRF(x, y ) d A , if R is the triangular region enclosed by the triangle of vertices ( 0 , O), ( 1 , O), and (0, 1) and F(x, y ) = ~ ~ ~ i+ xy2j.

11. Let F(t) be continuous for a 5 t 5 b and let q be a constant vector. Prove: a) ~ : q ~. ( t ) d= t q .jab~(t)dt

b) 1,bq x F ( t ) d t = q x ~ : F ( t ) d t

For functions of one variable the chain rule

at once gives the rule for change of variable in a definite integral:

Here f ( x ) is assumed to be continuous at least for x l 5 x 5 x2, x = X ( U )is defined for u l 5 u 5 u2 and has a continuous derivative, with x l = x ( u l ) , x2 = x(u2), and f [ x ( u ) ] is continuous for u 1 5 u 5 u2. Proof. If F ( x ) is an indefinite integral of f ( x ) , then

l:

,

f ( x )d x = F ( x 2 ) - F ( x l ) .

But F [ x ( u ) ] is then an indefinite integral of f [ x ( u ) ] % , for (4.59) gives

when x is expressed in terms of u . Thus the integral on the right of (4.60) is Since this is the same as the value of the left-hand side of (4.60), the rule is established. It is worth noting that the emphasis in (4.60) is on the function x ( u ) rather than on its inverse u = u ( x ) . Such an inverse will exist only when x is a steadily increasing function of u or a steadily decreasing function of u . This is not required for (4.60). In fact, the function x ( u ) can take on values outside the interval x l 5 x 5 xz, as illustrated in Fig. 4.7. However, f [ x ( u ) ] must remain continuous for u I 5 u 5 uz.

.

Chapter 4 Integral Calculus of Functions of Several Variables

Figure 4.7

The substitutio~ix = x ( u ) in a definite integral.

Figure 4.8 integral.

Curvilinear coordinates for change of variables in a double

There is a formula analogous to (4.60) for double integrals:

11

11f

R,,

R,,,.

f ( x , y ~ d x d y=

I I

[ ~ ( u.I. . Y(" v)1 a(x' dudv. a(u9u )

(4.61)

Here the functions 7 X

=X(U, v),

y = Y(U.

V)

are assumed to be defined and have continuous derivatives in a region R,, of the uv-plane. The corresponding points ( x , y) lie in the region R,, of the xy-plane, and it is assumed that the inverse functions

are defined and continuous in R,,, so that the correspondence between R,, and R,, is one-to-one, as suggested in Fig. 4.8. The function f ( x , y ) is assumed to be continuous in R,,, so that f [ x ( u ,v ) , y ( u , v ) ] is continuous in R,,. Finally, it is assumed that the Jacobian

is either positive throughout R,, or negative throughout R,,. It should be noted that it is the absolute value of J that is used in (4.61).

237

238

Advanced Calculus, Fifth Edition A proof of (4.61) will be given in the next chapter, with the aid of line integrals. Here we discuss the significance of (4.61) and its applications. Equations (4.62) can be interpreted as an introduction of curvilinear coordinates in the xy-plane, as suggested in Fig. 4.8. The lines u = const and v = const in R,, form a system of curves like the parallels to the axes. It is natural to use them to cut the region R,, into elements of area AA for formation of the double integral. With such curvilinear elements the volume "beneath the surface z = f (x,y)" will still be approximated by f (x, y)AA, where AA denotes the area of one of the curvilinear elements. If AA can be expressed as a multiple k of AuAv and f is expressed in terms of u and v , one obtains a sum which approaches a double integral

as limit. The crucial question is thus the evaluation of the factor k. As (4.61) shows, one must prove that

The number k can also be interpreted as the ratio of an element of area AA,, in the xy-plane to the element AA,, = AuAv in the uv-plane. Thus one must show (cf. Section 2.7) that

.j

EXAMPLE 1 Let x=rcosO,

y=rsin8,

so that the curvilinear coordinates are polar coordinates. The element of area is approximately a rectangle with sides r A 8 and Ar, as shown in Fig. 4.9. Thus

Figure 4.9

Element of area in polar coordinates.

1

Chapter 4 Integral Calculus of Functions of Several Variables and one expects the formula

J J f ( x . Y ) ~ X =~ Y11f (r cos 0 , r sin 0)r do dr.

(4.64)

R,,,

R,,

Now the Jacobian J in this case is

Thus (4.64)is correct.The region RrOcan be pictured in an re-plane or, more simply, can be described by inequalities such as the following:

which can be read off the figure in the xy-plane. From (4.65),one finds

so that the integral has been reduced to an iterated integral in r and 0. It may be necessary to decompose the region RrO into several parts and obtain the integral as a sum of iterated integrals of the form given. For some problems it is simpler to integrate in the order do, d r ; the region RrO must be described by inequalities

and the integral becomes

.

lblz:

f ( r cos 8 , r sin B)r dB dr.

EXAMPLE 2

Let it be required to evaluate

where R,, is the parallelogram shown in Fig. 4.10. The sides of R,, are straight lines having equations of form

for appropriate choices of cl ,c2. It is therefore natural to introduce as new coordinates

The region R,, then corresponds to the rectangle 1 5 u 5 4, -2 5 v 5 1. The correspondence is clearly one-to-one. The Jacobian is

239

240

Advanced Calculus, Fifth Edition

;i

Y

v

U

X

Figure 4.10

Curvilinear coordinates u = x

+ y , v = x - 2y. 4

Hence

It should be noted that the limits of integration for the uv-integral are determined from the figure and are not directly related to the limits that would be assigned to the corresponding iterated integral in the xy-plane. The fundamental formula (4.61) generalizes to triple integrals and multiple integrals of any order. Thus

///

f(x, y. z)dxdydz =

R,,:

// /

/

v 3

R,,,,",

1

a ( x ' Y ' z ) dudvdw, (4.66)

F ( ~W)~a(.. v, w,

where F(u, v, w) = f [x(u, v, w),y(u,v , w),z(u,v , w ) ] , under corresponding assumptions. Two important special cases are those of cylindrical coordinates:

SSS

f(x, y,z)dxdydz =

R,,-

/I/

F(r, 8. z)r d r dB dz,

R,,,.

(4.67) F(r, 8, z) = f (r cos 8. r sin 8, z), and spherical coordinates:

.

These are discussed in Problems 8 and 9. Remark. In order to determine the region R,, for (4.61)and, in particular, to verify that the correspondence between R,, and R,, is one-to-one a variety of techniques are available. In (4.62) or (4.63), one can set u = const and plot the resulting level curves of u in R,,. The same can be done for v. If these level curves have the property that a curve u = cl meets a curve v = cz in at most one point in R,,, then

3i

Chapter 4 Integral Calculus of Functions of Several Variables the correspondence must be one-to-one. From the level curves, one can follow the variation of u and v on the boundary of R,, and thereby determine the boundary of R,,. It can be shown that if R,, and R,, are each bounded by a single closed curve, as in Fig. 4.8, if the correspondence between (x, y) and (u, v)is one-to-one on these boundary curves, and J # 0 in R,,, then the correspondence is necessarily one-toone in all of R,, and R,,. For a further discussion of this point, the reader is referred to Section 5.14.Actually, the conditions that the correspondence be one-to-one and can that J # 0 are not vital for the theorem. It is shown in Section 5.14 that (4.61) be written in a different form that covers the more general cases.

PROBLEMS 1. Evaluate with the aid of the substitution indicated: a) J; (1 - x ~ )dx, ~ x/ = ~ sin B I dx, x = u 2 - 1 b, 1 . m

d)

sinx - c o s x ) dx, Son/4 x cosx(x I +xcosx

t = 1 +xcosx

2. Prove the formula JuI

as a special case of (4.60). 3. a) Prove that (4.60) remains valid for improper integrals, that is, if f (x) is continuous for x l 5 x < x2, X(U)is defined and has a continuous derivative for ul 5 u < u2, with X ( U I= ) X I , lim x(u) = x2, and f [x(u)] is continuous for ul. 5 u < u2. [Hint: U--*U? Use the fact that (4.60) holds with u2 and x2 replaced by uo and xo = x(uo), U I < uo < u2. Then let uo approach u2. One concludes that if either side of the equation has a limit, then the other side has a limit also and the limits are equal. Note that x2 or u2 or both may be m.] b) Evaluate 4 sinh 1 dx by setting u = $.

Jr

x-

,

Evaluate SobO(l- tanhx)dx by setting u = tanhx. 4. Evaluate the following integrals with the aid of the substitution suggested: a) SIR,, (1 - x ~ - ~dx~dy, ) where R,, is theregionx2fy2 5 1, usingx = r cos0, y = r sin8; C)

IfRY? x

+Y

dx dy, where R is the region I 5 x 4 2.0 5 y 5 x, using x = r cos 0. y = r sin@; c) SSR,,(x- y)2sin2(x + y)dx dy, where R,, is the parallelogram with successive vertices (n, 0), (2n,n), (n, 2n),(0, n), using u = x y, v = x y;

b)

-

+

SSR $ dx dy, where R is the trapezoidal region bounded by the lines x + y = 1, x + y = 2 in the first quadrant, using u = 1 + x + y, v = x - y; e) SJRJ5x2 + 2xy + 2y2dx dy over the region R bounded by the ellipse 5x2 + 2xy +

d)

(1-

)2

+

2y2 = 1, using x = u

+ v , y = -2u + v .

241

1

242

f

Advanced Calculus, Fifth Edition

5. Verify that the transformation u = ex cos y, v = ex sin y defines a one-to-one mapping of the rectangle R,,: 0 9 x 5 1,0 5 y 5 n / 2 onto a region of the uv-plane and express as an iterated integral in u, v the integral e2, dxdy. I e4, C O S ~y sin2 y R,, 6. Verify that the transformation

SS

+

defines a one-to-one mapping of the square 0 Ix 5 1,0 5 y 5 1 onto a region of the uv-plane. Express the integral

11;/r4

. ?I

+ y4dxdy

- 6x2y2

R,, over the square as an iterated integral in u and v. 7. Transform the integrals given, using the substitutions indicated: log(1

a)

+ x2 + y2)dydx, x = u + v,

y =u -v

id

+

b) {dj;_fxx d+dy dx, x = U , y = u v 8. Let R,, be the square 0 5 u 5 1 , 0 5 v 5 1. Show that the given equations define a one-to-one mapping of Ruuonto R,, and graph R,,: b) x = ueU, y = eU a) x = u + u 2 , y = eU d) x = 5u - u2 + v2, y = 5v + lOuv C) x = 2 u - v 2 , y = v + u v i 9. Verify the correctness of (4.67) as a special case of (4.66). Show the geometric meaning of the volume element r Ar A0 Az. 10. Verify the correctness of (4.68) as a special case of (4.66). Show the geometric meaniqg of the volume element p2 sin @ApA@A8. 11. Transform to cylindrical coordinates but do not evaluate: a) I D R , , ,x2y dx dy dz, where Rxyzis the region x2 y2 _( 1,0 Iz 5 1;

+

1

I+x+y

2

b) lolo lo ( X - y2)dzdy dx. 12. Transform to spherical coordinates but do not evaluate: a) f&,, x2y dx dy dz, where R,,, is the sphere: x2 y2

+ + z2 5 a2;

b)

,!I

.IF7

1

Im(x2

+ Y' + z2)dzdydx.

I

1

4.7 ARCLENGTH AND SURFACE AREA In elementary calculus it is shown that a curve y = f (x), a 5 x 5 b, has length

and that if the curve is given parametrically by equations x = x ( t ) , y = y(t) for t, 5 r 5 t2, then it has length

3

4

Chapter 4 Integral Calculus of Functions of Several Variables Furthermore, a curve x = x(t), y = y (t), z = z(t) in space has length

The length is here defined as a limit of the lengths of inscribed polygons; the functions are assumed to have continuous derivatives over the intervals concerned. The main question here is the generalization of (4.69) and (4.71) to the area of surfaces in space. For a surface z = f (x, y) it will be seen that the area is given by

This parallels (4.69). A surface in space can be represented parametrically by epuations

where u and v vary in a region R,, of the uv-plane. The area of the surface (4.73) is given by

where

A complete justification of (4.72) and (4.74) turns out to involve a very delicate analysis, much more difficult than that of arc length. In particular, one cannot define surface area S simply as the limit of the areas of inscribed polyhedra. For a full discussion the reader is referred to P. Franklin's Treatise on Advanced Calculus, pages 371-378 (New York: Dover, 1964). Here we give an intuitive discussion of why (4.72) and (4.74) are to be expected. We first note that the arc length s can be defined in a somewhat different way, using tangents rather than an inscribed polygon, as suggested in Fig. 4.11. Thus for a curve y = f (x), one subdivides the interval a _( x 5 b as for integrating f (x). At a point x: between x,-, and x,, one draws the corresponding tangent line to the curve:

243

244

Advanced Calculus, Fifth Edition Y

Figure 4-11

Definition of arc length.

and denotes by A; T the length of the segment of this line between xi.-, and xi. It is natural to expect that

as n becomes infinite and max Aix approaches 0. Now if u,*is the angle of inclination of Ai T, so that f '(xi*)= tan a;,one has

i

The sum C A, T thus becomes

If f l ( x ) is continuous for a 5 x expression

_(

b, this sum approaches as limit the desired

Now let a surface z = f ( x , y) be given, where f ( x , y) is defined and has continuous partial derivatives in a domain D. To find the area of the part of the surface above a bounded closed region R,,, contained in D, we subdivide R,, as in Fig. 4.3. Let (x:, y:) be a point of the ith rectangle; we then construct the tangent plane to the surface at the corresponding point:

Let Ai S*be the area of the part of this tangent plane above the ith rectangle in the xy-plane. Thus AiS* is the area of a certain parallelogram whose projection on the

Chapter 4 Integral Calculus of Functions of Several Variables

Figure 4.12

Definition of surface area.

-

xy-plane is a rectangle of area A i A , as in Fig. 4.12. Now let n* be the normal vector to the surface at the point of tangency:

Then it follows readily from geometry that A,S* = sec y,*A,A,

(4.79)

where y,* is the angle between n* and k . This is analogous to (4.76) (cf. Problem 5). Now from (4.78), COS yr* =

1

J

n*.k -- '"*I '

so that

all derivatives being evaluated at (x:, y:). Paralleling the procedure for arc length given earlier, it is natural to expect the surface area S to be obtained as the limit of a sum A;S* =

sec y , ' ~ i ~

245

246

Advanced Calculus, Fifth Edition

Figure 4.13

b ~3tqs.'

f~~

I

Curvilinear coordinates on a surface.

1

as the number n of subdivisions approaches oo, while the maximum diagonal approaches 0. This limit is precisely the double integral

s=

//secydxdy R,,

=

//,/q2)2+

(g)2cixdy

74

R,,

as desired.

I

Sulfaces in parametric form.

,.

+

x = x(u, v),

The parametric equations

y = y(u, v),

z

= z(u, v),

(u, v ) in R,,,

(4.80)

of a surface can be regarded as a mapping of R,, onto a ' ' C U N ~ ~region" in space, as suggested in Fig. 4.13. The lines u = const, v = const on the surface determine curvilinear coordinates on the sulface. When v = const, (4.80) can be regarded as parametric equations of a curve; the tangent vector to such a curve is the vector

in accordance with Section 2.13. Similarly, a curve u = const has tangent vector

The lines u = const, v = const in R,, can be used to subdivide R,, as in the formation of the double integral. Each rectangle, of area AA = AuAv, of this subdivision corresponds to a curved area element on the surface. To first approximation this area element is a parallelogram with sides IP, 1 Au and IP21Av, for P I can be interpreted as a velocity vector in terms of time u, so that

where s l is distance along the line v = const chosen. For a small change Au in the time u the distance moved is approximately d s , = IPl,lAu. Similarly, the other side of the parallelogram is approximately IP21 Av. The vectors P I and P2 are here evaluated at one comer of the parallelogram, as suggested in Fig. 4.13. The parallelogram has

Chapter 4 Integral Calculus of Functions of Several Variables then as sides segments representing the vectors P I A u and P 2 A v . Its area is thus given by

If we grant that the sum of the areas of these parallelograms over the surface approaches the area of the surface as limit as the number of subdivisions of R,, is increased, as in forming the double integral, then we conclude that the surface area is given by

By means of the vector identity

(see Problem 12(a) following Section 1.5) this can be written in the form

Comparison of (4.81), (4.82), and (4.75) shows that one has

Thus (4.85) reduces at once to the formula desired:

It should be noted that if the point ( x , y , z ) traces a general curve on the surface, the differential d r = dxi

+dyj +d z k

can be expressed as follows:

The element of arc on such a curve is defined by d s 2 = d X 2+ d y 2 + d Z 2= ldr12 = (PId u Expanding the last product, one finds

,:,,

r

+ P2 d v ) . (PIdu + P2 d v ) .

247

248

!i:3iwr

Advanced Calculus, Fifth Edition

I

Accordingly, by (4.86), ds2 = E du2

+ 2~ d u d v + G dv2.

(4.89)

This shows the significance of E, F, G for the geometry on the surface.

4

PROBLEMS 1. Find the length of the circumference of a circle a) using the parametric representation b) using the parametric representation 1-t2 x=u1 t2' 2. Find the area of the surface of a sphere a) using the equation

+

y =a-

2t 1+t2'

z = Ja2--, b) using the parametric equations

a'

c

)

3. Find the surface area of the surface having the given parametric equations: a) x = ( b + a c o s v ) c o ~ u , y = (b+acosv)sinu, z = a s i n v , 0 5 u p 217, 0 p v 5 2 x , where a and b are constants, 0 < a < b (torus). b) x = u cos v, y = u sin v , z = u2 sin2v, 0 5 u p 1, 0 p v 5 x / 2 (portion of saddle surface z = 2xy). 4. Using the parametric equations of Problem 2(b), set up a double integral for the area of a portion of the earth's surface bounded by two parallels of latitude and two meridians of longitude. Apply this to find the area of the United States of America, approximating this by the "rectangle" between parallels 30°N and 47"N and meridians 75"W and 122"W. Take the radius of the earth to be 4000 miles. 5. Let a parallelogranl be given in space whose sides represent the vectors a and b. Let c be a unit vector perpendicular to a plane C. a) 'Show that a x b .c equals plus or minus the area of the projection of the parallelogram on C. b) Show that this can also be written as S cos y , where S is the area of the parallelogram and y is the angle between a x b and c. C) Show that one has

s=

1

1

Jw,

where S,, , S,, , S,, are the areas of the projections of the parallelogram on the yz-plane, a-plane, q-plane. 6. A surface of revolution is obtained by rotating a curve z = f (x), y = 0 in the xz-plane about the z-axis. a) Show that this surface has the equation z = f (r) in cylindrical coordinates. b) Show that the area of the surface is

1

Chapter 4 Integral Calculus of Functions of Several Variables 7. Prove (4.74) by assuming that the surface in question can also be represented as z = f (x, y) and hence has an area given by (4.72). The parametric equations (4.73) are to be regarded as a transformation of variables in the double integral (4.72), as in Section 4.6.

8. Show that (4.74) reduces to

when u , v are curvilinear coordinates in a plane area R,,. 9. Prove that if a surface z = f ( x , y) is given in implicit form F ( x , y , z) = 0,then the surface area (4.72) becomes

Since the discontinuities of functions of several variables can be much more complicated than those of functions of one variable, the discussion of improper multiple integrals is not as simple as that for ordinary definite integrals. However, in principle, the analysis of Section 4.1 carries over to multiple integrals. The definition of boundedness of a function of several variables is the same as that of a function of one variable. Thus the function log (x2 y2) is bounded for 1 _< x 5 2, 1 5 y ( 2 but is not bounded for 0 < x 2 y2 ( I. If a function f ( x , y) is defined and continuous throughout a bounded closed region R such as (4.31), except for a finite number of points of discontinuity, and if f is bounded, then the double integral

+

+

continues to exist as a limit of a sum, for the same reason as for functions of one variable. The discontinuities can even take the form of whole curves, finite in number and composed of "smooth curves," for the similar reason that these together form a set of zero area. An important case of this is when f ( x , y) is continuous and bounded only in a bounded domain D, nothing being known about values of f on the boundary of D (assumed to consist of such smooth curves). In such a case the integral

continues to exist as a limit of a sum, provided that one evaluates f only where it is given, that is, inside D. The same result is obtained if one assigns f arbitrary values, for example, 0, on the boundary of D.

249

250

Advanced Calculus, Fifth Edition

EXAMPLE

The integral

// I sin

dx d y ,

where D is the square domain 0 < x < 1 , 0 < y < 1, exists, even though the function is badly discontinuous on the y axis, since I sin :I 5 1 in D. For any such bounded function f , one can approximate the integral over the whole region arbitrarily closely by the integral over a smaller region, avoiding the discontinuities. For if R is split into two regions R I , R2, overlapping only on boundary points, then

4

Further, if 1 f 1 p M in R, then

//

i

?

f b , y)dx dy 5 A4 . &.

(4.90)

R2

where A2 is the area of R2, as follows from (4.47). If A2 is sufficiently small, the integral over R1 will approximate the integral over R as closely as desired. True improper integrals arise, as for functions of one variable, when f (x, y ) is unbounded in R . The most important case of this is when f ( x , y ) is defined and continuous but unbounded in a bounded domain D, with no information given about values o f f on the boundary of D. In this case the limit of the sum C f (x, y ) A A will fail to exist because f is unbounded. The integral of f over D is termed improper and is assigned a value by the limit process

provided that the limit exists. Here R denotes a closed region contained, with its boundary, in D, as suggested in Fig. 4.14. The limit process is understood as follows: The limit exists and has value K if, given t > 0, a particular region R1 can be found such that I

I

for all regions R containing R 1 and lying in D. What amounts to a special case of the preceding is the case of a function having apoint discontinuity, that is, a function continuous in a domain D except at a single point P of D. It would be natural in this case to isolate the trouble at P by integrating up to a small circle of radius h about P and then letting h approach 0 (Fig. 4.15). This definition of the improper integral is equivalent to the preceding, provided that

'

Chapter 4 Integral Calculus of Functions of Several Variables

Figure 4.14

The limit process R -+ D. ,'CQ'/

Figure 4.15

,

3-lr>1

'

Integral improper at a point P.

f (x, y ) has the same sign (+or -) near P. This is the most frequent case. Thus the integral

+

where R is the circle x 2 y2 _< 1, is improper because of a point discontinuity at the origin. The limit process gives, in polar coordinates,

Thus if p < 2, the integral converges to the value 2 n / ( 2 - p ) ; if p > 2, the integral diverges. For p = 2, one obtains a logarithm and the integral again diverges. It should be noted that the critical value here is p = 2, as compared with p = 1 for single integrals.

251

252

Advanced Calculus, Fifth Edition

A second type of improper integral, generalizing the definite integrals with infinite limits, is an integral -?

J/f(X,Y)dXdY R

where R is an unbounded closed region. Here one obtains a value by a limit process just like that of (4.9 1). The most important case of this is that of a function continuous outside and on a circle x 2 y2 = a2. If f ( x , y ) is of one sign, the integral over this region R can be defined as the limit

+

where Rk is the region a2 5 x 2

has the value lim

k+ w

+ y2 ( k2. Thus the improper integral

lk12r

r: dB r d r = lim 2 n

4

k2-P - a 2 - ~

2

k-+m

-

~'

which equals 2 ~ a ~ --~2),/ for ( ~p > 2. For p 5 2 the integral diverges. While the emphasis here has been on double integrals, the statements hold with minor changes [affecting in particular the critical value of p for the integral (4.92)] for triple and other multiple integrals. For further discussion of this topic, see Section 6.26.

4

J

PROBLEMS 1. One way of evaluating the error integral

&

is to use the equations

(lm dm law e-.'

dx)l =

ePx' dx

3

e-"*dx

Z

e-y2dy =

JoWLm

7

e-x--v-

,

dx dy

and to evaluate the double integral by polar coordinates. Carry out this evaluation, showing that the integral equals f i ;also discuss the significance of the above equations in terms of the limit definitions of the improper integrals. ' 2. Show that the integral

//

log ~

m

d dy x

R

+

converges, where R is the region x2 Y 2 5 1 , and find its value. This can be interpreted as minus the logarithmic potential, at the origin, of a uniform mass distribution over the circle.

44

Chapter 4 Integral Calculus of Functions of Several Variables

3. a) Show that the integral

over the spherical region x2 + y2 + z2 5 1 converges for p <: 3 and find its value. For p = 1 this is the Newtonianpotential of a uniform mass distribution over the solid sphere, evaluated at the origin. b) For the integral of part (a), let R be the exterior region x2 + y2 + z2 2 1. Show that the integral converges for p 7 3 and find its value. 4. Test for convergence or divergence: a ) ~ ~ R G 21d ~ d y , o v e r t h e s q u a r e I x l < 1,Iyl

1;

i

SIR log(" y2) d x d y over the circle x2 + Y 2 9 1 ; JFT? c ) ISRlog ( x 2 + y 2 ) dx d y over the region x2 + y2 2 1 ;

b)

X2

d)

tr l

+

IJR

x2

+

2f '' d x d y over the region x2 + y 2 5 1 ;

e) J ] ] , log ( x 2 + y2

+ z 2 ) d xd y d z over the solid x2 + y2 + z2 5 1.

A definite integral

of a continuous function f (x, t) has a value that depends on the choice o f t , so that one,can write

lb

f ( x , t)dx = F(t).

(4.93)

One calls such an expression an integral depending on a parameter, and t is termed the parameter. Thus

is an integral depending on the parameter k; this example happens to be a complete elliptic integral (Section 4.2). If an integral depending on a parameter can be evaluated in terms of familiar functions, it becomes simply an explicit function of one variable. Thus, for example,

However, it can easily happen, as the preceding elliptic integral illustrates, that the integral cannot be expressed in terms of familiar functions. In such a case the function of the parameter is nevertheless well defined. It can be evaluated as accurately as desired for each particular parameter value and then tabulated; precisely this has

254

Advanced Calculus, Fifth Edition

rsjo

i

been done for the preceding elliptic integral, and tables thereof are easily accessible (Section 4.2). The question to be studied here is the evaluation of the derivative of a function F ( t ) defined by an integral as in Eq. (4.93):

LEIBNITZ'S RULE Let f (x, r) be continuous and have a continuous derivative af / a t in a domain of the xt-plane that includes the rectangle a 5 x p b, t, 5 t p t2. Then for t , 5 t 5 t2,

In other words, differentiation and integration can be interchanged; for example,

Here both sides can be fully evaluated so that the result can be checked.

Proof of Leibnitz's Rule.

Let

Since af l a t is continuous, one concludes from the theorem of Section 4.3 that g ( t ) is continuous for tl 5 t 5 t ~Now . for tl 5 t 3 5 t2,

by the theorem referred to one can interchange the order of integration:

where F ( t ) is defined by (4.93).If we now let r3 be simply a variable t , we have

Both sides can now be differentiated with respect to t. By the fundamental theorem ( 4 .lo), one obtains

Thus the rule is proved.

f.

Chapter 4 Integral Calculus of Functio* of Several Variables The notion of integral depending on a parameter extends at once to multiple integrals, and Leibnitz's Rule also generalizes. Thus d" a1 2 1 2 m d x d y = 1 2 1

xa logx

24-

+ ya logy d x dy.

It can be extended to improper integrals, but with complications; this is discussed in . . Chapter 6. One has at times to consider expressions like (4.93) in which t#%k\ts of integration a and b themselves depend on the parameter t . For example, one might have

Here again, one has a method for finding the derivative in terms of an integral.

THEOREM Let f (x, r) satisfy the condition stated previously for Leibnitz's Rule. In addition, let a(t) and b(t) be defined and have continuous derivatives for tl 5 t 5 t2. Then for tl 5 t 5 t2, f ( x , t > d x = f [b(t),tlbt(t)- f [a(t), t]af(t)+

t)dx.

(4.95) \

ht

Thus, for example,

Proof. Let u = b(t), v = a(?), w = t , so that the integral F(t) can be written 'as follows: F(t) =

/Cu f

(x, w)dx = G ( u , v , w),

where u , v , w all depend on t . Hence by the chain rule,

It will be seen that the three terms here correspond to the three terms on the right of (4.95). Indeed, one has

by the fundamental theorem (4.10). Since u = b(t), d u l d t = b'(t) and

The second term is accounted for similarly, the minus sign appearing because

255

256

Advanced Calculus. Fifth Edition Finally,

?

l"

'-a -- a aw

t

f( x , w ) d x =

aw

by Leibnitz's Rule. Since w = t , d w / d t = 1 and the third term is accounted for.

PROBLEMS 1. Obtain the indicated derivatives in the form of integrals: a)

$ 1: -dx

b)

ad i Jl 2 (1 -x2,,,z dx

C)

& 1; log (xu) dx

d)

&7

2. Obtain the indicated derivatives: a) 1; t2dt C)

d"

slnx 12, x--y dx

&

b) ai d Jlt 2 sin (x2)dx

$ ~ ? l o g ( l+ x 2 ) d x

d) x

+ j txa n x e-r2

4

dt

3. Prove the following: d

a) b) C)

cvsa

dx = 1% Ssina 1% $1: usinuxdx =O;

Coscr + a

a(;;

& /:'

-[sin

log (COSa + a )

+cos a log (sin a+a)]; J_ I

e-'2~'

dx = 2 y e - ~ *- e - ~ 4- Zy :J x'e-"'~' dx.

1;

4. a) Evaluate xn log x dx by differentiating both sides of the equation 1 with respect to n (n > - 1). b) Evaluate

1:

~ " e - "dx~ by repeated differentiation of

yy,),,

jr jr

1; xn dx

=

e-"" dx (a > 0).

(x2 by repeated differentiation of x2:y2. [In (b) and (c) the improper integrals are of a type to which Leibnitz's Rule is applicable, as is shown in Chapter 6. The result of (a) can be explicitly verified.] 5. Leibnitz's Rule extends to indefinite integrals in the form:

e) Evaluate

There is still an arbitrary constant in the equation-because we are evaluating an indejnite integral. Thus from the equation

1

one deduces that Jxetxdx=eu(5-$)+~I.

a) By differentiating n times, prove that dx /(x2+ar

-

(-1)n-1

an-1

(n-l)!aan-l 8''

slnax

b) ~ r o v e j x ~ c o s a x d=x ;i;;"(7)

+ ~ , =n4 , 8 , 12, ....

i

,

.T

4

Chapter 4

1

f ( x , t ) d x = F(x, t) equivalent to the statement

c) Let

Integral Calculus of Functions of Several Variables

+ C, so that a F / a x

257

= f ( x , t). Show that Eq. (a) is t

a 2 -~ a 2 ~

ax at

at a x '

6. It is known that C O S ~ 1-Jc-7 do = 2n 1-acose a J F 2 ' where a is a constant, 0 < a < 1. (This can be established as in elementary calculus with the aid of the substitution t = tan (8/2).) Use this result to prove that

2n

12"

log (1 - a cos 0) d e = 2n log

l + d n 2 .

[Hint: Call the left-hand side of the new equation g(a), find gl(a), and integrate to find g(a) = 2n log (1 0) C . Use continuity of g for a = 0 and g(0) = 0 to find C.] 7. Consider a 1-dimensional fluid motion, the flow taking place along the x axis. Let v = V(X, t) be the velocity at position x at time t, so that if x is the coordinate of a fluid particle at time t, one has d x l d t = v. If f (x, t ) is any scalar associated with the flow (velocity, acceleration, density,. . . ), one can study the variation of f following the flow with the aid of the Stokes derivative: af dx af - - - -~t ax d t at [see Problem 12 following Section 2.81. A piece of the fluid occupying an interval a0 5 x ( bo when t = 0 will occupy an interval a(t) 5 x 5 b(t) at time t , where = = v(b, t). The integral v(a, t),

+

+

+-

%

F(t) =

1;;

f (x, t ) d x

is then an integral of f over a definite piece of the fluid, whose position varies with time; i f f is density, this is the mass of the piece. Show that

This is generalized to arbitrary 3-dimensional flows in Section 5.15. 8. Let f ( a ) be continuous for 0 5 a 5 2n. Let

for r < 1, r and 8 being polar coordinates. Show that u is harmonic for r < 1. This is the Poisson integral formula. [Hint: Write w = 1 r 2 - 2r cos (8 - a ) and v(r, 8 , a ) = (1 - r2)w-I. Use Leibnitz's Rule to conclude that v 2 u = (2n)-' f ( a ) ~ ~ v d ~ , where v 2 v = v,, r-2vee r-'v, as in Eq. (2.138) in Section 2.17. Show that v, = -2rw-' - (1 - r ' ) ~ - ~ wetc. , and finally

+

+

+

1:

Multiply both sides by r 2w3, insert the proper expressions for w, w,, . . . on the right and collect terms in powers of r (r6, r5,. . .) to verify that ~'w'v'v = 0 and hence v 2 u = 0.1

258

Advanced Calculus, Fifth Edition

!, ~a%tki

*4.10 UNIFORM CONTINUITY 8 EXISTENCE OF THE RIEMANNINTEGRAL In this section and the following one we provide additional real variable theory and prove the existence of single and double integrals. The theory can be regarded as a continuation of that developed in Section 2.23; hence we continue naming theorems by letters (Theorem K, and so on) and refer to similar theorems in Section 2.23. DEFINITION Let f be a mapping of the set G in E N into EM.Then f is said to be uniformly continuous if for each r > 0 there is a 6 > 0 such that whenever P , Q are in G and d ( P , Q ) < 6, one has d ( f ( P ) , f ( Q ) ) < c. a A uniformly continuous mapping is clearly continuous at every point. However, 6 now depends only on r and not on the point. For example, y = f ( x ) = x 2 , 0 5 x < co,is a continuous mapping of the set G : 0 5 x < co into E l , but this mapping is not uniformly continuous. Thus let us take r = 1 and ask how close x l , x2 must t = be so that ( f ( x l ) - f(x2)l < 1. If Ix2 - x l I = c , then ( f ( x l ) - f(x2)l = ~ x -xi1 Ixl x211x2 - x 11 = clxl x21. If this is to be less than 1, then we must have

+

+

+

But we can choose x l , x2 as close together as we wish while J x l x21 is arbitrarily large. Thus there is no 6 > 0 such that c < 6 implies ) f ( x l ) - f (x2)1 < 1 . We can illustrate this numerically by trying to calculate y = x 2 with error less than 0.001 by approximating x by a nearby value with small error. For x = 1 we approximate by 1.0004 and get

d'i

y = (1.0004)' = 1.00080016, tr)

'

p=u

which is within the allowed error. For x = 10, use of the same accuracy in x leads us to 10.0004 and y = 100.0080016, so that the error in y is now greater than 0.001. The larger the x , the smaller the error in x allowed to ensure the same accuracy in y; the 6 depends on E and x . THEOREM K Let f be a continuous mapping of a bounded closed set G in EN into E M .Then f is uniformly continuous.

Proof; Suppose f is not uniformly continuous. Then for some c > 0 and every 6 > 0 we can find a pair of points P , Q in G such that d ( P , Q ) < 6 but d (f ( P ) , f ( Q ) ) > r . In particular, we take 6 successively equal to 112, 114, . . . , 1/2", . . . to obtain sequences Pn, Q , such that d(Pn. Qn) <

1

2"

and d ( f (Pn). f(Q.1)

2 c.

3 ?

By Theorems E and F of Section 2.23 we can choose a subsequence P,, converging to Po in G . Then

Chapter 4 Integral Calculus of Functions of Several Variables Since both sequences on the right have limit 0, we conclude that Q,, -+ Po also. Hence f (Pn,) f (PO), f (en,)+ f (PO).But

0<

i d(f (pn,), f (Qn,))

i d(f (Pn,), f (PO))+ d(f (PO), f( e n l ) ) .

As k -+ GO, both terms on the right have limit 0. That is a contradiction. Hence f is uniformly continuous.

THEOREM L Let y = f (x)be a continuous real function of the real variable x for a 5 x ( b. Then the Riemann integral

exists. Proof. We first observe that the interval a 5 x J b is a bounded closed set (Problem 4 following Section 2.23). Hence by Theorem K , f is uniformly continuous. As in Section 4.1, we now consider subdivisions of the interval: a = xo c x, i. . . < x, = b. We let h be the mesh of the subdivision, the maximum of A1x = X I - XO, . . . , Anx = x, - x,-,. We must show that there is a unique number c such that for each t > 0 there is a 6 > 0 such that

for all subdivisions of mesh h less than 6, no matter how xt, . . . , x,* are chosen in 5 xT 5 x, for i = 1, . . . , n. the successive subintervals: To this end, we first observe that by Theorem J of Section 2.23, f has an absolute maximum B and minimum B, so that A 5 f (x)5 B for a 5 x 5 b. Hence

for all sums C f (x:)A,x. Thus the sums form a bounded set of real numbers. In particular, we consider Es, the set of sums arising only from those subdivisions of mesh h less than 6, where 6 > 0. We let ,. , ~ ~ s = g l b E ~ /,f 3 s = l ~ b E s . Thus

for all subdivisions of mesh h less than 6.

259

260

Advanced Calculus, Fifth Edition

4 id

By uniform continuity we now choose q > 0 so that

whenever x', x u are on the interval and Jx' - x"l < q. We let 6 = q / 2 and consider two sums in E g .The two sums may come from different subdivisions of the interval. However, by combining the points of the two subdivisions we obtain a new subdivision of mesh h less than 6 and can write each sum in terms of the new subdivision. If a point 2 is a new subdivision point between xi-1 and xi, we write

and can proceed similarly for all new subdivision points. Thus both sums now have the form, in terms of the new subdivision,

However, the points x: , x:* are no longer necessarily between xi- 1 and x, .We can say that in any case, x: andx:* are at most 26 = q apart. Accordingly, ( f (x:)- f (x:*)( < c / [ 2 ( b - a ) ] and hence

Hence any two numbers in Eg are less than e / 2 apart and therefore /?g - a6 5 c/2. We now choose the sequence 6, = 2-" (n = 1,2,. . .) converging to 0 and let sn be the corresponding sequence as,, of greatest lower bounds of E6,,. Since Ec+, is a subset of E6,, (every subdivision of mesh less than 2-"-' is a subdivision of mesh less than 2 7 s, 5 s, for all n , so that s, is a monotone increasing sequence; this sequence is also bounded, since s, 5 B(b - a ) for all n. Therefore, by Theorem A of Section 2.23, s, converges to some number c, so that Is, - cl < €12 for n sufficiently large. But for n sufficiently large, 6, is less than 6 as above, so that lag,, - Bg,,I 5 c / 2 and every sum s in E6,,satisfies 1s - snI ( €12. Therefore for n sufficiently large, E e < - + - = E

2

2

for every sum s in Es,,; that is, for 6 sufficiently small and 0 < h < 6, we have Is - cl < E for every sum s obtained from a subdivision of mesh h. Therefore the theorem is proved.

9

Chapter 4 Integral Calculus of Functions of Several Variables

We now prove two theorems that show the existence of the double integral and justify its representation as an iterated integral. We confine attention to regions G of form a Ix i b, Y I ( X ) I Y i Y ~ ( x ) . THEOREM M Let f ( x , y) be continuous on the set G : a I x Ib, yl ( x ) 5 y 5 y2(x), where y ~ ( x and ) y 2 ( x ) are continuous for a 5 x Ib and y l ( x ) 5 y2(x) for a I.x Ib. Then

is continuous for a F: x Ib. Proof. We first consider the case when yl ( x ) and y2(x) are unequal constant functions so that

the function f is now continuous on the bounded closed set a Ix 5 b, c Iy 5 d . Hence it is uniformly continuous: Given E > 0, we choose 6 > 0 so that If ( x ,, y2) f ( ~ 2 ~, 2 1 1< ~ l ( d c ) when & ( X I ,Y I ) ,( ~ 2y2)) , < 6. Then for 1x1 - x2I < 6 ,

since d ( ( x l ,y ) , ( x 2 ,y ) ) = Ixl - x21 < 8 . Therefore g is continuous for a 5 x 5 b. For the general case (4.96) we make the substitution, for fixed x ,

As in Section 4.6. we obtain

+

. -)is As a composite function of continuous functions, the integrand f ( x , y l ( x ) continuous for a 5 x 5 b, 0 5 u 5 1. The integral now has the form (4.97), with c = 0, d = 1. Therefore by the case considered previously it defines a continuous function of x for a 5 x 5 b; multiplication by the continuous function y2(x) - yl ( x ) produces a continuous function g ( x ) . Therefore the proof is complete.

261

262

Advanced Calculus, Fifth Edition

*

I

THEOREM N Let f (x , y) be continuous on the set G : a 5 x 5 b, y (x) 5 y 5 y2(x), where yl(x) and y2(x) are continuous for a 5 x 5 b and yl(x)< y2(x) for a < x < b. Then the double integral of f over G exists and equals the iterated integral

Proof. As in Section 4.3, we subdivide the region by parallels to the axes: Lines x = X, = const for a = xo < xl < . . . < x, = b, lines y = y, = const for yo < y~ < . . . < y,. Here yo must be less than or equal to the minimum of yl(x), and y, must be greater than or equal to the maximum of y2(x). These lines form rectangles, and we let h be the mesh of the subdivision, the largest diagonal of the rectangles. By Theorem M the iterated integral (4.98)exists and has a value c. We let g(x) denote the inner integral, as in (4.96), so that g(x) is continuous. Now

by the Mean Value theorem, with xi-, 5 xf 5 xi and A i x 1, . . . , n. Similarly,

xi -

for i =

Here we use the values y,, to subdivide the interval yl(x:) 5 y 5 y2(x:), letting y,I be the first of the values yo, yl, . . . to exceed yl(x:) and y,,,,-1 be the last to fall below Y~(x:). We also write

Y ~ O= YI(x:),

Ytm, = Y~(x:)

(see Fig. 4.16). We write

13

Figure 4.16

Formation of sum (4.101).

,

%

?

Chapter 4 Integral Calculus of Functions of Several Variables Also the values y,: , y:2,

263

. , . satisfy

If we combine (4.99) and (4. loo), we obtain

On the right-hand side we have a sum like that for the double integral; on the left-hand side we have a constant c. It follows that as the mesh h + 0, the sums on the right-hand side have limit c. However, the sum in (4.101) includes terms for rectangles not wholly contained in G. This is clear from Fig. 4.16, where the top and bottom rectangles extend beyond G. We show below that for given r, a 6 > 0 can be found so that for 0 < h < 6 the absolute value of the sum of such terms is less than r/2. Therefore we can write, for O
(4.102)

Granting this point, we now consider an arbitrary sum for the double integral, using the same subdivision by lines parallel to the axes; this allows us to vary the choice of the points at which f is evaluated in each rectangle. We obtain a number

where the sum is as in (4.101), but we exclude the contribution from rectangles extending beyond G, as in the first term on the right of (4.102). Hence we can write c-S =

x[

f (x:.

y3) - f (x:*. y::)]AixAik y

+ term of absolute value < r/2.

Now G is a bounded closed set. Hence f is uniformly continuous on G. Therefore we can choose 6 so small that the values of f at points less than S apart differ by less than 6/(2A), where A is the area of G. For subdivisions of mesh h, with h < 6 and 6 thus restricted (and restricted as earlier), we have from (4.104)

For here C C AixAiky is the total area of the rectangles contained in G, and this area is at most A. Accordingly, by definition of the double integral,

//

f (x, y)dA = c =

% I

G

ir) as asserted. It remains to show that for 6 sufficiently small the rectangles extending beyond G contribute a sum of absolute value less than 612 to the whole sum (4.101). Now f (x, y) is continuous on G, and therefore If (x, y)l 5 K for some constant K , for all (x, y) in G. This follows from Theorem J of Section 2.23. Accordingly, the

Advanced Calculus, Fifth Edition

Figure 4.17

Contribution of area due to boundary curve.

absolute value of the sum we are considering is at most K times the sum of the areas of the rectangles. Thus it suffices to show that the total area of the rectangles can be made arbitrarily small by choosing h sufficiently small. But, as Fig. 4.17 shows, the extruding rectangles in the interval x,-, 5 x 5 xi provide along the upper boundary a total area less than (max Y ~ ( x-) min y2(x))Aix

+ 2hAix.

(4.105)

By uniform continuity, for h sufficiently small, max y2(x) - min yz(x) is less than a given I] > 0, and if h is also than I], the whole expression (4.105) is less than 31]A;x, so that the sum of the areas from the upper boundary is at most 31](b- a ) . If I ] = ~ / [ 1 2 ( b- a ) ] ,the upper boundary contributes less than €14 to the area of extruding rectangles. Similarly, the lower one contributes less than c/4 to this area. Hence for h sufficiently small the total unwanted area is at most €12. Thus our conclusion follows. Remarks. The argument about the extruding rectangles shows why, in finding the double integral as a limit, one can include or exclude such terms. For h sufficiently small their contribution is as small as desired. Roughly stated, the boundary curves here have zero area. There are bounded regions whose boundary sets do not have zero area, and for these our limit process for the double integral leads to difficulty. To handle such problems and, in general, to provide a more satisfactory theory of integration, it is better to use the Lebesgue integral, which is developed in many advanced texts including that of Rudin listed at the end of the chapter.

PROBLEMS 1. a) Prove the rule for vectors in

vN :

.-. - r

llul - lvll 5 lu - vl. b) Prove: The function f (x) = 1x1 is a continuous mapping of EN into E l . Is it uniformly continuous? 2. Determine whether the function is uniformly continuous: a) y = e X , O(x11; b) y = logx, 0 < x 5 1; C) y =logx, 1 < x < 00; d) y =sinx, -m < x < m. 3. Prove: If f ( x ) is defined for a c x < b, f '(x) is continuous for a < x < b and f '(x) is bounded, If '(x)( ( K, then f is uniformly continuous for a < x < b.

1

Chapter 4

Integral Calculus of Functions of Several Variables

4. Let f (x, y ) have continuous partial derivatives in the domain D in the xy-plane. Further let l V f 1 5 K in D, where K is a constant. In each of the following cases, determine whether this implies that f is uniformly continuous: a) D is the domain x2 + y2 < 1. [Hint: If s is distance along a line segment from (xl , y l ) to (x2, y2),then f has directional derivative d f l d s = V f . u along the line segment, where u is an appropriate unit vector.] b) DisthedomainIxl < 1,lyl < 1,excludingthepoints(x,0)forO~x< 1. - - -

5. Show that if f and f ' are continuous for a 5 x 5 b and I ff(x)l 5 K = const for a 5 x 5 b, then for each subdivision of mesh less than 6 = c / [ 2 K ( b- a ) ] ,each sum C f (x:)A,x differs from jabf (x) d x by less than 6.

Suggested References Courant, Richard J., and Fritz John, Introduction to Calculus and Analysis, vol. 1, New York: Interscience, 1965, and vol. 2, New York: John Wiley & Sons, 1965. Franklin, Philip, A Treatise on Advanced Calculus. New York: Dover, 1964. Goursat, Edouard, A Course in Mathematical Analysis, vol. 1, trans. by E. R. Hedrick. New York: Dover Publications, 1970. Henrici, Peter K., Elements of Numerical Analysis. New York: John Wiley and Sons, Inc., 1964. Kaplan, Wilfred, and Donald J. Lewis, Calculus and Linear Algebra, 2 vols. New York: John Wiley and Sons, Inc., 1970-1971. Ralston, Anthony, and Rabinowitz, Philip, First Course in Numerical Analysis, 2nd ed. New York: McGraw-Hill, 1978. Rudin, Walter, Principles of Mathematical Analysis, 3rd ed. New York: McGraw-Hill, 1976. Scarborough,James B., Numerical Mathematical Analysis, 6th ed. Baltimore: Johns Hopkins Press, 1966. von K h i n , Theodore, and Maurice A. Biot, Mathematical Methods in Engineering. New York: McGraw-Hill, 1940. Whittaker, E. T., and G. N. Watson, Modern Analysis, 4th ed. Cambridge: Cambridge University Press, 1940. Widder, David V., Advanced Calculus, 2nd ed. New York: Dover Publications, 1989.

265

Vector Integral Calculus

PART I. TWO-DIMENSIONAL THEORY

The topic of this chapter is line and surface integrals. It will be seen that these can both be regarded as integrals of vectors and that the principal theorems can be most simply stated in terms of vectors; hence the title "vector integral calculus." A familiar line integral is that of arc length: lcd s . The subscript C indicates that one is measuring the length of a curve C, as in Fig. 5.1. If C is given in parametric form x = x ( t ) , y = y ( t ) , the line integral reduces to the ordinary definite integral:

If the curve C represents a wire whose density (mass per unit length) varies along C , then the wire has a total mass

c

where f ( x , y) is the density at the point (x, y) of the wire. The new integral can be expressed in terms of a parameter as previously or can be thought of simply as a limit of a sum

268

Advanced Calculus, Fifth Edition

Figure 5.1

Line integral.

Figure 5.2

Work = J, F . dr.

Here the curve has been subdivided into n pieces of lengths A l s , Azs, . . . , A,s, and the point (x:, y:) lies on the ith piece. The limit is taken as n becomes infinite, while the maximum A,s approaches 0. A third example of a line integral is that of work.If a particle moves from one end of C to the other under the influence of a force F, the work done by this force is defined as

where F+ denotes the component of F on the tangent T in the direction of motion. This integral can be thought of as a limit of a sum as previously. However, another interpretation is possible. We first remark that the work doneby a constant force F in moving a particle from A S B on the line segment A B is F . AB; for this scalar product is equal and hence to the product to IF1 . cos or 1AB 1 . o being the angle between F and of force component in direction of motion by the distance moved. Now the motion of the particle along C can be thought of as the sum of many small displacements along line segments, as suggested in Fig. 5.2. If these displacements are denoted by Air, A z r , . . . , A,,r, the work done would be approximated by a sum of form

-

3.

Chapter 5 Vector Integral Calculus where Fi is the force acting for the ith displacement. The limiting form of this is again equal to the line integral FTd s , but because of the way the limit is obtained, we can also write it as

/

F .dr.

C

One can thus write =SF-dr.

work= C

C

If the displacement vector A r and force F are expressed in components,

F = Fxi + F,j,

A r = Axi

+ Ayj,

the element of work F . A r becomes

F . A r = F,Ax

+ F,Ay.

The total amount of work done is then approximated by a sum of form

The limiting

ib;m of this is a sum of two integrals:

The first integral represents the work done by the x-component of the force; the second integral represents the work done by the y-component of the force. It thus appears that one has three types of line integrals to consider, namely, the tY Pes

which are limits of sums

f ( ~ Y) 1 As,

C P(x. y ) Ax.

Q(x, y) Ay.

The foregoing gives the basis for the theory of line integrals in the plane. A very slight extension of these ideas leads to line integrals in space:

Surface integrals appear as a natural generalization, with the surface area element da replacing the arc element ds:

269

270

Advanced Calculus, Fifth Edition There are corresponding component integrals //f(x,~.z)dxdy.

. .

//f(x,y,z)dydz, S

S

and a vector surface integral

/I

F . d~ = J/(F. n) d ~ ,

S

S

where d o = n d a is the "area element vector," n being- a unit normal vector to the surface. It will be seen that the basic theorems-those of Green, Gauss, and Stokesconcern the relations between line, surface, and volume (triple) integrals. These correspond to fundamental physical relations between such quantities as flux, circulation, divergence, and curl. The applications will be considered at the end of the chapter. 8

5.2

LINEINTEGRALS

b

,

~

<

* $ ,

IN THE PLANE

We now state in precise form the definitions outlined in the preceding section. By a smooth curve C in the xy-plane will be meant a curve representable in the form:

where x and y are continuous and have continuous derivatives for h 5 t 5 k. The curve C can be assigned a direction, which will usually be that of increasing t . If A denotes the point [#(h),I,b(h)]and B denotes the point [#(k),I,b(k)],then C can be thought of as the path of a point moving continuously from A to B. This path may cross itself, as for the curve C1of Fig. 5.3. If the initial point A and terminal point B coincide, C is termed a closed curve; if, in addition, (x, y ) moves from A to B = A without retracing any other point, C is called a simple closed curve (curve C2 of Fig. 5.3).

Figure 5.3

Paths of integration.

* 4

Chapter 5 Vector Integral Calculus

Figure 5.4

Definition of line integral.

Let C be a smooth curve as previously, with positive direction that of increasing t . Let f ( x , y ) be a function defined at least when ( x , y ) is on C . The line integral Jc f ( x , y ) d x is defined as a limit:

The limit refers to a subdivision of C as indicated in Fig. 5.4. The successive subdivision points are A: (xo, yo), ( x l, y . . . , B : ( x , , y,). These correspond to parameter values: h = to < t l < . . < t,, = k . The point ( x ; , y:) is some point of C between y i - , ) and ( x i , y i ) ; that is, (x*, y;) corresponds to a parameter value ti*, where ti-, 5 t; 5 t i . A i x denotes the difference xi - x i - l . The limit is taken as n becomes infinite and the largest A i t approaches 0, where A i t = ti - t i - l . Similarly,

-

/

(5.3)

f ( x , y ) d y = l i m x f ( x ~Y, : ) A , Y ,

C

where A , y = y, - Y , - ~ . The significance of these definitions is guaranteed by the following basic theorems:

I. If f ( x , y ) is continuous on C , then

1

f

(x,y)dx

and

11. If f ( x , y ) is continuous on C , then

Sf

(x,y)dy

exist.

271

272

Advanced Calculus, Fifth Edition 1

Formulas (5.4) and (5.5) reduce the integrals to ordinary definite integrals and are thus essential for computation of particular integrals. Thus let C be the path x = 1 + t , y = t 2 , 0 5 t 4 1 , directed with increasing t . Then

It is logically easier to prove I1 first, for I is an immediate consequence of 11. To prove 11, one notes that the sum C f (x:, y:)A,x can be written as

Now A i x = xi - x i - ] = @'(tf*)Airby the Mean Value theorem. Hence the sum can be written as

where F ( t ) = f [ # ( t ) , $ ( t ) ] and tf and tf* are both between ti-] and t i . It is easily shown [see CLA, Section 12-25] that this sum approaches as limit the integral

as required. Formula (5.5) is proved in the same way. We remark that the value of a line integral on C does not depend on the particular parametrization of C, but only on the order in which the points of C are traced. (See Problem 5 after Section 5.3.) In many applications the path C is not itself smooth but is composed of a finite number of arcs, each of which is smooth. Thus C might be a broken line. In this case, C is termed piecewise smooth. The line integral along C is simply, by definition, the sum of the integrals along the pieces. One verifies at once that (5.2), (2.3),and the theorems I and I1 continue to hold. In ( 5 . 4 ) and (5.5) the functions @'(r)and @ ' ( t ) will have jump discontinuities, which will not interfere with the existence of the integral (cf. Section 4.1). Throughout this book all paths of integration for line integrals will be piecewise smooth unless otherwise speciJied. If the curve C is represented in the form

then one can regard x itself as parameter, replacing equations

x=x,

y=g(x),

r ; that is, C is given by the

a i x ~ b

273

Chapter 5 Vector Integral Calculus in terms of the parameter x . If the direction of C is that of increasing x , (5.4) and (5.5) become

The ordinary definite integral fab y d x , where y = g ( x ) , is a special case of (5.6). Similarly, if C is represented in the form

and the direction of C is that of increasing y, then

.

In most applications the line integrals appear as a combination,

which is abbreviated as follows:

the brackets being used only when necessary. In the formulas thus far the direction of C has been that of increasing parameter. If the opposite direction is chosen, upper and lower limits are reversed on all integrals. Thus (5.4) becomes

The line integral is therefore multiplied by - 1. Often it is convenient to specify the path by its equations in some form and to indicate the direction by using the initial and terminal points as lower and upper limits: LBpdX+edy C

or

/

( x ? .v ? )

(XI.",)

P ~ X Q + ~ Y .

1

274

Advanced Calculus, Fifth Edition

Figure 5.5

19

Example '1.

It will be seen later that under certain conditions, one needs only prescribe initial and terminal points:

EXAMPLE 1

To evaluate

where C is the semicircle y = parametrically:

JZi shown in Fig. 5.5, one can represent C

and the integral becomes a?

3lr (cos3t - sin3t ) cost dt = 8 ' One can use x as parameter, and the integral becomes

this is clearly in a more awkward form for integration. The substitution x = cost brings one back to the parametric form. One can use y as parameter but has then to split the integral into two parts, from (1, 0) to (0, 1) and from (0, 1) to (- 1,O):

Note that x = ,/part. EXAMPLE 2

on the first part of the path and x = -J-

on the second

Let C be the parabolic arc y = x 2 from (0,O) to (- 1, 1). Then

Chapter 5 Vector Integral Calculus

(a)

(b)

.

f h=

Figure 5.6 (a) Positive direction and (b) negative direction on a simple closed path.

Figure 5.7

Example 3. 1

If C is a closed curve, then there is no need to specify initial and terminal point, though the direction must be indicated. If C is a simple closed curve (traced just once), then one need only specify which of the two possible directions is chosen. The notations

refer to the two cases of Figs. 5.6(a) and 5.6(b). The counterclockwise arrow refers to what is roughly a counterclockwise direction on C; this will be termed the positive direction (as for angular measure); the clockwise direction will be called the negative direction. It should be noted that the direction can be specified by reference to the unit tangent vector T in the direction of integration and the unit normal vector n that points to the outside of the region bounded by C; for the positive direction, n is 90" behind T, as in Fig. 5.6(a); for the negative direction, n is 90" ahead of T as in Fig. 5.6(b). EXAMPLE 3

To evaluate #y2dx +x2dy. C

where C is the triangle with vertices (1, O), (1, I), (0, 0), shown in Fig. 5.7, one has to compute three integrals. The first is the integral from (0, 0) to (1, 0); along this

. 275

276

Advanced Calculus, Fifth Edition

1

path, y = 0 and, if x is the parameter, d y = 0. Hence the first integral is 0. The second integral is that from (1,O) to (1, 1); if y is used as parameter, this reduces to

since d x = 0. For the third integral, from (1, 1 ) to (0, 0), x can be used as parameter, so that the integral is

since dy = dx. Thus finally

#

C

5.3

2

'

2

1

4

ydx+x dy=O+l--=-. 3 3

INTEGRALS WITH RESPECT TO ARCLENGTH rn BASICPROPERTIES OF LINEINTEGRALS

d

For a smooth or piecewise smooth path C , as in the preceding section, arc length s is well defined. Thus s can be defined as the distance traversed from the initial point (t = h) up to a general t : $

If the curve C is directed with increasing t , then s also increases in the direction of motion, going from 0 up to the length L of C. Let C be subdivided as in Fig. 5.4 and let A i s denote the increment in s from ti..., to ti,that is, the distance moved in this interval. One then makes the definition

If f is continuous on C , this integral will exist and can be evaluated as follows:

This is proved in the same way as (5.4) and ( 5 . 3 , with the aid of the formula

One can in principle use s itself as the parameter on the curve C ; if this is done, x and y become functions of s : x = x(s), y = y(s). The point [x(s),y(s)] is then the position of the moving point after a distance s has been traversed. In this case,

Chapter 5 Vector Integral Calculus (5.11) reduces to a definite integral with respect to s:

277 8r

If x is used as parameter, one has

r

there is an analogous formula for y . The basic combination

1

= 7.b ~ d x + a d y

h

\

referred to above can be written as an integral with respect to s as follows:

C

C

, '1 .+ i . b3. where cr is the angle between the positive x-axis and a tangent vector in direction of increasing s. For, as shown in Fig. 5.8 (cf. Section 2,13), one has

dx = COSCY, ds

dy = sincr ds

and

The basic properties of the line integral are analogous to those of ordinary definite integrals; the properties to be listed here can in fact be demonstrated by reducing the line integral to a definite integral by use of a parameter: L;(Pdx+Qdy)+ CI

Figure 5.8

(Pdx+Qdy)=

c2

Element of arc.

(Pdx+Qdy), C7

(5.18)

278

Advanced Calculus, Fifth Edition where C3 is the combined path: from A l to

via C1and from A2 to A3 via C2;

where C' denotes C traced in the reverse direction;

K

S

(Pdx

C

+ Q dy) = S(KP)dx + (KQ)dy,

-

K = const;

(5.21) m

C

S

ds = L = length of C ;

C

if

If

(x, y)I 5 M on C, then

if C is a simple closed curve, as in Fig. 5.6(a) then

= area enclosed by C

These theorems all hold under the assumptions that the paths are piecewise smooth and that the functions being integrated are continuous. All except the formula (5.24) follow at once from appropriate parametric representations; (5.24) will be proved in Section 5.5.

PROBLEMS 1. Evaluate the following integrals along the straight-line paths joining the end points: (232) 2

a) S(0,O) Y d x ,

b) );$.',

Y dx,

c)

x dy.

c3,'))

\ 2. Evaluate the following line integrals: a)

hO,-,)y

2

dx

+ x 2 d y , where C is the semicircle x = dm;

C

b)

fgbg) y d x + x d y , where C is the parabola y = x 2 ;

C

c)

Y-,

where C is the curve x = eos3 t , y = sin3r. 0 5 t 5 f.

C

(Hint: Set u = tan3 t in the integral for t . )

Chapter 5 Vector Integral Calculus

3. Evaluate the following line integrals: a) $,y2 dx + xy dy, where C is the square with vertices (1, I), (-1, I), (-1, (1, -1); b) $cy dx - x dy, where C is the circle x2 + y2 = 1 (cf. (5.24)); c) $,x2y2 dx - xy3 dy, where C is the triangle with vertices (0, O), (1, O), (1, 1). 4. Evaluate the following line integrals: a) $C(x2 - y2) ds, where C is the clrcle x2 y2 = 4; r 4 i i . (v . J A b) x ds, where C is the line y = x;

+

h:,;; c) ~;d,i; ds. where C is the parabola y = xi.

-I),

*?

C

x)9

C

5. Let a path (5.1) be given and let a change of parameter be made by an equation t = g(t), (Y 5 t 5 B, where gl(r) is continuous and positive in the interval and g(a) = h , g(B) = k. As in(5.4) the lineintegral Sf (x, y) dx on thepathx = 4(g(t)), y = $(g(s)) is given by

Show that this equals the integral in (5.4), so that such a change of parameter does not affect the value of the line integral. 6. (Numerical evaluation of line integrals) If the parametric equations of the path C are explicitly known, Eqs. (5.4) and (5.5) reduce the evaluation of line integrals to a problem in ordinary definite integrals, to which the methods of Section 4.1 apply. One can in any case evaluate the integral directly from its definition as limit of a sum. Thus

/

Pdx

+ Qdy

C

-

~ ( P ( X : , y:)A,x 1=1

+ Q(x:.y:)A;y).

The points (xf , y:) can be chosen as the subdivision points (xi-1, Y;-~,)or as (xi, y;); one can also use a trapezoidal rule to obtain the sum: " 1 1 ( I [ ~ ( X i - l , Y;-I) P(xi9y;)]A;x 5[Q(x;-l. Y;-1) Q(x;. Y;)]A;Y]. (a) i=l For example,

c

+

+

+

where C is the straight line segment from (0, 0) to (2, 2) and the subdivision points are (O,O), (1, 11, (2,2). Let the functions P and Q be given by the following table at the points A . . . S. A B C D E F G H I J K L M N O S

1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 0 3 8 5 3 0 5 2 8 5 0 1 2 7 3 4 1 2 3 4 2 4 6 8 3 6 9 2 4 8 2 6

280

Advanced Calculus, Fifth Edition Evaluate the integral J P dx + Q dy approximately by the trapezoidal rule, Eq. (a) above, on the following broken line paths: (a) ABFG (b)AFGKH (c) ABCDHLSONMIEA (d) AFJNMIJFA (e) ABFEAEFBA. 7. Show that if f (x, y) > 0 on C, the integral Sc f (x, y)ds can be interpreted as the area of the cylindrical surface 0 5 z 5 f (x, y), (x, y ) on C .

The functions P(x, y) and Q(x, y) appearing above can be regarded as the components of a vector u:

u = P(x, y)i + Q(x, y)j,

u, = P(x, Y),

u, = Q(x, Y).

(5.25)

The line integral

has then a simple vector interpretation as follows:

where U T denotes the tangential component of u, that is, the component of u in the direction of the unit tangent vector T in the direction of increasing s . For as Fig. 5.9 shows (cf. Section 2.13), T has components d x l d s , dylds: dx T = -i ds

+ dy -j ds

= cosai

+ sinaj.

(5.27)

Hence UT

=u.T= PCOS~ Qsina, +

so that

by (5.15).

In the case in which u = P i + Q j is a force field, the integral J U T d s = [P d x + Q dy represents the work done by this force in moving a particle from

Figure 5.9

Unit tangent vector.

I

281

Chapter 5 Vector Integral Calculus one end of C to the other. For the work done is defined in mechanics as "force times distance" or, more precisely, as follows: work =

S

(5.30)

(tangential force component) d s ,

and u r is precisely this tangential force component. The line integral [P dx Q dy can be defined directly as a vector integral as follows:

+

where [u.dr=

n+oo lirn

CU(X:,Y;).A,I .

maxA,s+O 1=1

n C

I

,I

.v'

2

and A,r = Aixi + A,yj. Equation (5.17) can then be written in the vector form:

'

'

-

.'>f

If C is represented in terms of a parameter t , then

If r is the position vector of a particle of mass m moving on C and u is the force t: 1s applied, then by Newton's Second Law,

Hence, if Ivl = v ,

=

.

lk

dt

(lm 2 y2)

dl.

One thus concludes that

that is, the work done equals the gain in kinetic energy. This is a basic law of mechanics.

282

Advanced Calculus, Fifth Edition

dy ds

Figure 5.10

The line integral way, namely, as

Tangent and normal.

1, P dx + Q dy can be interpreted as a vector integral in another /v-nds =/vnds.

*-

=

C

C

where v is the vector Qi - Pj and n is the unit normal vector 90" behind T, so that v, is the normal component of v. One has then (cf. Fig. 5.10)

since the vector product a x k of a vector a in the xy-plane with k is a vector b having the same magnitude as a and 90" behind a. Accordingly,

Thus one has

as asserted. It should be noted that for u = Pi

+ Qj,one has then

4

The following theorem and its generalizations are fundamental in the theory of line integrals.

Chapter 5 Vector Integral Calculus

GREEN'S THEOREM Let D be a domain of the xy-plane and let C be a piecewise smooth simple closed curve in D whose interior is also in D. Let P(x, y) and Q(x, y) be functions defined and continuous and having continuous first partial derivatives in D. Then

where R is the closed region bounded by C The theorem will be proved first for the case in which R is representable in both of the forms: ~ 1 . a9 ~ ~ . a 5 x 5 b, c

5 Y 5 d,

fi(x) 5 y 5 f2(~),

(5.41)

g~(y5 ) x

(5.42)

5 g2(y),

as in Fig. 5.1 1. The double integral

I ,

R

by (5.41) can be written as an iterated integral:

One can now integrate out:

Figure 5.11

Special region for Green's theorem.

283

284

Advanced Calculus, Fifth Edition

Figure 5.12

Decomposition of region into special regions.

2

In the same way, J J R d x dy can be written as an iterated integral, with the aid of (5.42), and one concludes that

,

Adding the two double integrals, one finds:

I/ (g

$

-E)dxdy =fPdx+

R

Qdy.

The theorem is thus proved for the special type of region R. Suppose next that R is not itself of this form but can be decomposed into a finite number of such regions: R 1 ,R2, . . . , R, by suitable lines or arcs, as in Fig. 5.12. Let C1,C2,. . . , C, denote the corresponding boundaries. Then Eq. (5.40) can be applied to each region separately. Adding, one obtains the equation:

f f =/I(%-%)

$(pdx+ady)+ Cl

f3

C

(

)+.-

ay

1

c,,

C2

ax

'

dxdy+...+/J(

RI

)dxdy.

3

R,,

But the sum of the integrals on the left is just

For the integrals along the added arcs are taken once in each direction and hence cancel each other; the remaining integrals add up to precisely the integral around C in the positive direction. The integrals on the right add up to

i

Chapter 5 Vector Integral Calculus

285

The proof thus far covers all regions R of interest in practical problems. To prove the theorem for the most general region R , it is necessary to approximate this region by those of the special type just considered and then to use a limiting process. For details, see the book by Kellogg cited at the end of the chapter.

+ y2 = 1. Then

EXAMPLE 1

Let C be the circle: x2

EXAMPLE 2

Let C be the ellipse x2

1 3,

v

-8

+ 4~~ = 4. Then

where A is the area of R. Since the ellipse has semi-axes a = 2, b = 1, the area is nab = 2n, and the value of the line integral is 4n. EXAMPLE 3 Let C be the ellipse 4x2 applicable to the integral

+ y2 = 4. Then Green's theorem is not

since P and Q fail to be continuous at the origin. EXAMPLE 4 Then

Let C be the square with vertices (1, I), (1, -I), (- 1, -I), (- 1, 1).

. .

where (X,7 ) is the centroid of the square. Hence the value of the line integral is 0. o Vector interpretation of Green's theorem. If u = P(x, y)i previously,

+ Q(x,y)j, then, as I

where U T = u . T is the tangential component of u. The integrand in the right-hand member of Green's theorem is the z component of curl u; that is, a~ a p t curlz u = - - -. 1 ax ay i Hence Green's theorem states:

This result is a special case of Stokes's theorem, to be considered later.

286

Advanced Calculus, Fifth Edition 4

The integral $, P dx

+ Q dy can also be interpreted, as above, as the integral fv C

- n ds = fun ds, C

where v is the vector Qi - P j . The right-hand member of Green's theorem is then the double integral of div v. Thus one has, for an arbitrary vector field,

where n is the outer normal on C, as in Fig. 5.6(a). This result is the 2-dimensional form of Gauss's theorem, to be considered later.

Application to area. Equation (5.24) asserts that

This follows at once from Green's theorem since

If these two equal line integrals are averaged, another expression for area is obtained:

, ?

This can also be checked by Green's theorem. 0

PROBLEMS

+

+

1. If v = (x2 y2)i (2xy)j, evaluate 1, v~ ds for the following paths: a) from (0,O) to (1, 1) on the line y = x, b) from (0,O) to (1, 1) on the line y = x2; c) from (0,O) to (1, 1) on the broken line with corner at (1,O). 2. Evaluate lcun ds for the vector v given in Problem 1 on the paths (a), (b), and (c) of Problem 1, n being chosen as the normal 90" behind T. 3. The gravitational force near a point on the earth's surface is represented approximately by the vector -mgj, where the y axis points upwards. Show that the work done by this force on a body moving in a vertical plane from height h to height h2 along any path ,is equal to mg(hl - h2). 4. Show that the earth's gravitational potential U = -kMm/r is equal to the negative of the work done by the gravitational force F = - ( k ~ m / r ~ ) ( r / rin ) bringing the particle to its present position from infinite distance along the ray through the earth's center.

Chapter 5 Vector Integral Calculus

287

5. Evaluate by Green's theorem: a) $cay dx bx dy on any path;

+

b) $ex sin y dx

+ ex cosy dy around the rectangle with vertices (0, O), (1, O), (1, in),

(0,

+ y2 = 1; ds, where u = grad ( x ~and ~ )C is the circle x2 + y2 = 1; ds, where v = (x2 + y2)i - 2xyj, and C is the circle x2 + y2 = 1, n being the

c) $(2x3 - y3)dx + (x3 + y3) dy around the circle x2 d) $cur

e) $,u, outer normal; f ) $c [(x - 212

&

9) §c

C log

+ y2]ds, where C is the circle x2 + Y 2 = 1, n is the outer normal;

[ ( x - 2)'

+ ),2, ds, where C and n are as in (f);

+

h) $cf (x) dx g(y) dy on any path. 6. If r = xi yj is the position vector of an arbitrary point (x, y), show that

+

r,, ds = area enclosed by C, C

n being the outer normal to C 7. Check the answers to Problems 2(a), 3(a), (b), (c), and 4(a) following Section 5.3 by Green's theorem.

Let the functions P ( x , y ) and Q(x, y) be defined and continuous in a domain D.' Then the line integral j P dx Q d y is said to be independent of path in D if, for every pair of endpoints A and B in D, the value of the line integral

+

l B P d x + Qdy is the same for all paths C from A to B. The value of the integral will then in general depend on the choice of A and B, but not on the choice of the path joining them. Thus as in Fig. 5.13, the integrals on C , , C2, C3 have the same value.

Figure 5.13

Independence of path.

I

288

Advanced Calculus. Fifth Edition

Figure 5.14

Construction of function F such that dF = P dx

+ Qdy.

THEOREM I If the integral P d x + Q d y is independent of path in D, then there is a function F ( x , y ) defined in D such that

holds throughout D. Conversely, if a function F ( x , y ) can be found such that (5.46) holds in D , then j" P d x + Q d y is independent of path in D.

Proof. Suppose first that the integral is independent of path in D. Then choose a point (xo, yo) in D and let F ( x , y ) be defined as follows:

where the integral is taken on an arbitrary path in D joining (xO,yo) to ( x , y). Since the integral is independent of path, the integral in (5.47) does indeed depend only on the endpoint ( x , y ) and defines a function F ( x , y). It remains to show that a~la= x P(X, y), a ~ l a = y Q ( X ,y ) in D. For a particular ( x , y ) in D, choose ( X I , y ) so that xl # x and the line segment from ( x l ,y ) to ( x , y ) is in D , as shown in Fig. 5.14. Then, because of independence of path, one has

Here x l and y are thought of as fixed, while ( x , y ) may vary along the line segment. Thus y has been restricted to a constant value, and F ( x , y ) is being considered as a function of x near a particular choice of x . The first integral on the right is then independent of x , while the second can be integrated along the line segment. Hence for this fixed y ,

or, with the dummy variable x replaced by a t ,

F ( x , y ) = const

+

1:

P ( t , y)d't.

d

289

Chapter 5 Vector Integral Calculus ., . -The fundamental theorem of calculus (Section 4.1) now gives ' aF = P(x, y). ax <4<; Similarly, aF/ay = Q(x, y), and the first part of the theorem is proved. t Conversely, let (5.46) hold in D for some F. Then, in terms of a parameter t , ,,it

In other words, in simplified notation, ,-,

'

lB + P dx

B

Qdy =

dF = F ( B ) - F ( A ) .

i

(5.48)

?l

C

The value of the integral is simply the difference of the values of F at the two ends • and is therefore independent of the path C. Remark 1. Formula (5.48) is the analogue of the usual formula

lb

b

f (x) dx = F(x)I~= F(b) - F(a),

-

.,

I * p T

F1(x) = f (x)

for evaluating definite integrals and should be used whenever the function F can be found. Thus

Remark 2. Suppose aF/ax = P shows that

= 0 in D, aF/ay

i!

=Q

xt~

= 0 in D. Then (5.48)

that is, that F(B) = ' F ( A ) for every two points of D. Hence F (cf. Section 2.22).

= const in

D

Remark 3. If P and Q are given such that J P dx + Q dy is independent of path, then the function F such that d F = P dx Q dy is determined only up to an additive constant:

+

290

Advanced Calculus, Fifth Edition For if G is a second function such that aG/ax = P, aG/ay = Q, then (a/ax) ( G - F ) = P - P = O , ( a / a y ) ( G - F ) = Q - Q =O;hencebyRemark2,G-F is a constant, or G = F const. It often happens that P = aFli3.x and Q = aF/ay are known (perhaps in tabular form), while F is not; in this case, (5.49) permits evaluation of F at any point in D, once its value at (xo,yo) has been chosen. For (5.49) can be written:

+

as the replacement of the upper limit (x, y) by (xo, yo) shows. In (5.50), any convenient path of integration can be used, for example, one formed of lines parallel to the axes. This is very important in thermodynamics, in which the partial derivatives of such functions as internal energy, entropy, and free energy are the measured quantities, rather than the functions themselves (cf. Section 5.15).

Remarkl. It is to be emphasized that the function F(x , y) in (5.46)must be defined and continuous in a domain containing the path. A function such as arc tan (ylx) is "many-valued" and hence useless for the theorem unless it is unambiguously defined and has the required continuity; these requirements cannot be satisfied when C is the circular path x = cost, y = sin t , 0 5 t _( 2n. This point is considered further in connection with Example 2. THEOREM I1

If the integral

1P d x + Q dy is independent of path in D, then

on every simple closed path in D. Conversely, if (5.5 1) holds for every simple closed path in D, then P dx Q dy is independent of path in D.

1

+

Prooj Suppose first that the integral is independent of path. Let C be a simple closed path in D and divide C into arcs A B and B A by points A , B as in Fig. 5.15. Then

1

Because of independence of path, one can write this equation in the form

Chapter 5 Vector Integral Calculus

%

Figure 5.15

Proof of first part of theorem 11.

Figure 5.16

Proof of second part of theorem 11.

<

.

Conversely, suppose [P d x + Q d y = 0 on every simple closed path in D. Let A and B be two points of D and let C, and C2 be two paths from A to B as in Fig. 5.16. We must show that

or, if C; denotes C2 traced in the reverse direction, (Pdx CI

+ Q d y ) + 1.

Pdx

+ Q d y =0.

c;

If the paths C, and C; do not cross, together they form a simple closed path C, and

regardless of the direction of integration. If the paths cross a finite number of times, as in Fig. 5.16, a repetition of the same argument gives the proof; this reasoning covers, in particular, the case in which C1 and C2 are broken lines. To cover the most general case, we can now reason as follows. We consider the function (5.47), using as paths only broken lines from (xO,yo) to (x, y).

291

292

Advanced Calculus, Fifth Edition We have just shown that the value is independent of the path on such paths, so that F ( x , y ) is well defined. Furthermore, we deduce exactly as previously that aF/ax = P(x, y ) , a G / a y = Q ( x , y ) , using only broken line paths. Hence by Theorem I the line integral is independent of path for all types of paths in D. Remark. The first part of the preceding proof shows that, if [ P d x independent of path, then

+ Q d y is

on every closedpath C in D , regardless of whether or not C crosses itself. THEOREM I11 If P ( x , y ) and Q(x, y ) have continuous partial derivatives in D and [P d x Q d y is independent of path in D, then

+

Proof. By Theorem I, one has

and, since P and Q have continuous derivatives in D, aP a 2 -=-ay ayax

~ a2F _--

d

.,

a, --

a ~ a y

ax'

Simply connected domains. The converse of Theorem III does not hold without a further restriction, namely, that the domain D be simply connected. In plain terms, a domain is simply connected if it has no "holes." Thus the domains of Figs. 5.13 and 5.14 are not simply connected, but rather are, as one says, multiply connected. More precisely, D is simply connected if for every simple closed curve C in D the region R formed of C plus interior lies wholly in D (cf. Fig. 5.17). Examples of simply connected domains are the following: The interior of a circle, the interior of a square, a sector, a quadrant, and the whole xy-plane.

Figure 5.17

Simply connected domain.

-7

Chapter 5 Vector Integral Calculus

Figure 5.18

293

Doubly connected domain.

The annular region between two circles is not simply connected, nor is the interior of a circle minus the center. Thus the holes may reduce to single points. One can distinguish between types of multiply connected domains as follows: A domain with just one hole, as in Fig. 5.18, is doubly connected; one with two holes is triply connected; one with n - 1 holes is n-tuply connected. One can even have infinitely many holes, in which case one speaks of a domain that is infinitely multiply connected. After this preparation we can state the converse of Theorem 111: THEOREM IV Let P ( x , y ) and Q ( x , y ) have continuous derivatives in D and let D be simply connected. If

+

in D, then J P d x Q d y is independent of path in D. This theorem and its consequences are fundamental in physical applications. Proof. Suppose (5.53) to be satisfied. Then choose any simple closed curve C in D. By Green's theorem,

where R is the region formed of C plus interior. It should be remarked that Green's theorem is applicable because D is simply connected, so that R is in D and, in particular, a Q / a x and a P / a y are continuous in R. Since we have shown that

1

for every simple closed path C in D, it follows from Tbeorem I1 that P d x is independent of path in D.

1

Vector interpretation. Theorem I asserts that P d x precisely when P = a F / a x , Q = a F / d y , that is,

+Qd y

•

+ Q d y is independent of path

1

294

1

Advanced Calculus, Fifth Edition

I

Thus S u T ds is independent of path precisely when u is the gradient vector of some scalar F: u = V F. Equation (5.48) then states that

+

Theorem 111 asserts that if j P dx Q dy is independent of path, then aP/ay = aQ/ax. If u = Pi Qj, this is the statement that

+

curl u = 0,

yl

.

8

for

Since we know by Theorem I that u is a gradient, Theorem 111 is equivalent to the statement curl grad F = 0. This fundamental identity was pointed out in Chapter 3 [Eq. (3.31)]. Theorem IV then provides the converse: I f curl u = 0 in D, then u = grad F for some E provided that D is simply connected. The analogous statement for three dimensions will be proved in Section 5.13.

EXAMPLE 1 The integral

is independent of path in the whole plane, since

To find a function F whose gradient is Pi

+ Qj, set

using a brol%~j.line path, as in Fig. 5.19. Along the first part, dy = 0 and y = 0,so that the integral reduces to 0. Along the second part, dx = 0, and x is constant, so that

Chapter 5 Vector Integral Calculus

Figure 5.19

Example 1.

The general solution is then x2y3 (1,2) to (3, -2), one then writes

EXAMPLE 2

+ C . TO evaluate the integral on any path from

The integral

is independent of path in any simply connected domain D not containing the origin, for aP- aQ y2 - x 2 - -ay ax (x2 y2)2 L .

+

i

+

except at the origin. To find the function F whose differential is P d x Q dy, one can proceed as in Example 1, using as path a broken line from (1, 0) to (x, y). However, P d x Q d y is here a familiar differential (cf. Example 3, Section 2.8), namely, that of the polar coordinate angle 0:

+

(

:)=

d 0 = d arc tan -

-ydx+xdy xZ+y2

.

The arc tangent is rather to be avoided if possible, because one cannot in general restrict to principal values and also because of the awkwardness of arc tan oo for the y-axis. One would therefore write

However, in order for 0 to be a well-defined continuous function in a domain including the path (which in particular requires that 0 be single-valued), so that Theorem I can be applied, it is necessary to restrict to a simply connected domain not containing the origin. Two such domains are shown in Figs. 5.20(a) and (b). In Fig. 5.20(c) the path is the circle x2+y2 = 1, and the domain is doubly connected. Since x 2 y = 1 on the path,

+

295

X

I 296

Advanced Calculus, Fifth Edition

(a) jnde = R.

(b)

=

jn.

CC)

4 -5d x+

+ X ~ Y=

),2

Zz,

t d

Figure 5.20

Example 2.

here Green's theorem, which is not applicable to the given integral, could be applied after the line integral had been simplified by using the relation x2 y2 = I on the path; one can also regard the line integral as a sum of integrals from 8 = 0 to 8 = rr and from 8 = n to B = 2 n , giving rr n = 2n. A similar procedure can be followed in general, and one concludes that for any path C not through (0, O),

+

+

= total increase in 8 from A to B, C

as 8 varies continuously on the path C . The integral is not independent of path but depends on the number of times C goes around the origin. The integral

EXAMPLE 3

is independent of path in the doubly connected domain consisting of the xy-plane minus the origin. First of all,

but, as Example 2 shows, this would guarantee independence of path only in a simply' connected domain. However, for this example, one has additional information:

.

.

xdx+ydy = d log x2 y2

+

Jm-,

1

Chapter 5 Vector Integral Calculus

297

1

and the function log ,/= = log r is well defined (that is, single valued), with continuous derivatives except at the origin. Hence

for any path from A to B not through the origin-in particular, then, for the paths of Fig. 5.20(a) and (b). On the circle of Fig. 5.20(c) the integral is 0, in accordance with Theorem 11.

5.7 EXTENSION OF RESULTSTO MULTIPLY CONNECTED DOMAINS If the closed curve C of Green's theorem encloses a point at which a P / a y or aQ/ax fails to exist, then the theorem cannot be applied. The following theorem shows that even in this case the reduction of a properly chosen line integral to a double integral is possible. THEOREM Let P ( x , y) and Q ( x , y) be continuous and have continuous derivatives in a domain D of the plane. Let R be a closed region in D whose boundary consists of n distinct simple closed curves C 1 ,C 2 ,. . . , C,, where C1 includes C2,. . . , C,, in its interior. Then

In particular, if aQ/ax = a P / a y in D, then

To prove the theorem, one first introduces auxiliary arcs from C 1 to C2, C 1 to C 3 ,. . . , two to each, as illustrated in Fig. 5.21. These decompose the region R into smaller regions, each of which is simply connected; there are four subregions

Figure 5.21

Green's theorem for multiply connected domains.

1 1

I

298

Advanced Calculus, Fifth Edition in Fig. 5.21. If one integrates in a positive direction around the boundary of each subregion and then adds the results, one finds that the integrals along the auxiliary arcs cancel out, leaving just the integral around C1in the positive direction plus the integrals around CZ,C3, . . . , in the negative direction. On the other hand, the line integral around the boundary of each subregion can be expressed as a double integral

over the subregion by Green's theorem. Hence the sum of the line integrals is equal to the double integral over R. This gives (5.55); Eq. (5.55') then follows as a special case. If one denotes by BR the directed boundary of R, that is, the curves C, , CZ,. . . , Cn with the given directions, then Eqs. (5.55) and (5.55') can be written in the concise form:

It is to be noted that the correct direction of integration always keeps the region to the left; that is, the normal pointing away from the region is 90" behind the tangent. As an application of the theorem just proved, let us consider the case of a doubly connected region D with a hole which, for the sake of simplicity, we take to be a point A, as in Fig. 5.22. Let a P / a y = dQ/ax in D. What are the possible values of an integral

about a simple closed path in D? We assert that there are only two possible values, namely, the value 0 when C does not enclose the hole A (the curve Co of Fig. 5.22) and a value k, which is the same for all curves C enclosing the hole A. That the value is 0 for a curve such as Co follows from the fact that Co lies in a simply connected

Figure 5.22

Doubly connected domain (point A excluded).

'

299

Chapter 5 Vector Integral Calculus part of D, so that Theorem IV of the preceding section is applicable. It remains to show that for paths CI and C2 which both enclose A , one has

To prove this, let us assume first that C1and C2 do not meet and that, for example, C2 lies within C1.Then CI and C2 form the boundary of a region R to which (5.55') is applicable:

as was to be shown. If C1 and C2 do intersect each other, a sufficiently small circle C3 about A will meet neither of them. One then has

as before. Thus again

and (5.57) is completely proved.

EXAMPLE 1

The integral

of Example 2 of the preceding section is of the type considered, with the hole A at the origin. Hence the integral is 0 when C does not enclose the origin and has a certain value k when C does enclose the origin. To find k, we choose C to be the circle x2 y2 = 1 and find that

+

This is in agreement with the previous results. EXAMPLE 2

.

Example 3 of rhe preceding section, I

fx::;P'. C

,

-

~-'

Advanced Calculus, Fifth Edition illustrates a case in which the constant value'k is 0.Thus the integral is 0 on all simple closed paths not through the origin. EXAMPLE 3 To evaluate

+

around the ellipse x 2 3y2 = 1, one verifies that a P / a y = aQ/ax except at (0, O), where both functions are discontinuous. Then, since the ellipse is awkward for integration, one replaces it by the circle x2 y2 = 1, which also encloses the point of discontinuity. With the parametrization x = cos t , y = sin t , the integral on the circle is

+

(-sin3 t sin t - cost sin2t cos t) dt = Thus the integral on the given ellipse equals -n. The results (5.55) or (5.56) can be put in vector form. Thus ds in (5.43) and (5.44) (Section 5.5), one has

/

UT

dr =

fl

curl," d r d y ,

(5.58)

where T is the unit tangent vector in the direction of integration and n is 90" behind

T, so that n is an outer normal, that is, points away from R. Remark. It can happen that the path C passes through a discontinuity of P or Q. If the line integral is reduced to a definite integral f (t) dt in the parameter t, one has then the problem discussed in Section 4.1. If f (t) is discontinuous only at a finite number of points and is bounded, then the integral continues to exist; if f (t) is unbounded, the integral is improper and may or may not exist.

lub

PROBLEMS 1. Determine by inspection a function F ( x , y ) whose differential has the given value and integrate the corresponding line integral: a) dF = 2xy dx

+ x 2 d y , -ht,bf/2xy d x + x2d y , where C is the curve y = x i ;

yexY dx + xexYdy, where C is thecurve y = sin3x;

b) dF= yexY dx +xeXYd;,

e

C

e) d

F

= (12

q

+ y2)5

, /'e2"'0)xdxtydy c(l,o)

(x2 + Y2)3 '

whereCisthecurvex=efcort,y=etsint.

2. Test for independence of path and evaluate the following integrals: (3,4) d x x d y on the line y = 3x - 5; 41.4) b) A!,;) $ d x - 4 d y on the parabola y = 2 + x2. Y-

a)

x2

1

Chapter 5 Vector Integral Calculus

+ (x2 + 6y) dy on the circular arc y = m, -1 (t,?) d) j(o,o) sec2x tan y dx + sec2 y tan x dy on the curve y = 16x3/n2. c) $ j ( 2 x y - 1) dx

301

5 x 5 1;

3. Evaluate the following integrals: a) $[sin(xy) xy cos(xy)] dx x2 cos(xy) dy on the circle x2 + y2 = 1;

+

+

circle x2 + y2 = 4; b, f c) $y3dx - x 3 d y on thesquare 1x1 lyl = 1; y d x - ( x - 1)dy on the (X-,)~+)J~

+

+

+

+

7

.

.

C ' *

.,."*

d) $xy6 dx ( 3 6x) ~ dy on ~ the ellipse ~ ~x2 4y2 = 4. e) $(7x - 3y + 2) dx + (4y - 3x - 5) dy on the ellipse 2x2 + 3y2 = 1; on the circle x2 + = [Hint: First show that the f, denominator § 0 and cosy 1 by writing it as (ex - 112 + 2eX is 0 only for x (e'cosy-l)dx+eXsinydy e2r-2etcosy+ 1

y2

= (1 - cosy).] 4. Determine all values of the integral

1.

=

i'

! a.

.

on a path not passing through the origin. 5. On a path C: x = f (t), y = g(t), a 5 t 5 b not passing through the origin, for which f ( a ) = f (b), g(a) = g(b), the analysis of Section 5.6 shows that

equals n . 2 n , where n is the number of times C encircles the origin (counted positive or negative according as 0 has a net increase or decrease on the path); n is called the winding number of C. The value of n can be determined from a sketch of the path. Evaluate the line integral for the following paths: a) x = 5 + c o s 3 t , y = 8 + s i n 3 t , 0 5 t 5 271 b) x = c o s t + tsint, y = s i n t , O s t 5 2 n c) x = 2cos2t -cost, y = 2sin2t - sint, 0 5 t 5 2 n

Remark In this way, some difficult definite integrals are evaluated with ease. For example, b) integrates cos2 t/(l 2t sin t cost t2 sin2 t) from 0 to 2n.

+

+

6. Show that the following functions are independent of path in the xy-plane and evaluate them: a) 2xy dx (x2 - y2) d y

k:",:)

+ b) J$$) sin y dx + x cos y dy.

.s

7. Evaluate

around the square with vertices (f1, f 1) (note that aP/ay = aQ/ax). 8. Let D be a domain that has a finite number of "holes" at points A ! , A2, . . . , Ak, SO that D is (k

+ 1)-tuplyconnected; cf. Fig. 5.23. Let P and Q be continuous and have continuous

Advanced Calculus, Fifth Edition

Figure 5.23

1 3

Problem 8.

derivatives in D, with aP/ay = aQ/ax in D. Let C1 denote a circle about A1 in D, enclosing none of the other A's. Let C2 be chosen similarly for A2, and so on. Let

a) Show that if C is an arbitrary simple closed path in D enclosing Al , A 2 , . . . , Ak, then

b) Determine all possible values of the integral

d

lxl.,l~ ( ~ 2 ~ ~ 2 )

P d x + Qdy ?df.4 between two fixed points of D, if it is known that this integral has the value K for one particular path. 9. Let P and Q be continuous and have continuous derivatives, with aP/ay = aQ/ax, except at the points (4, O), (0, O), (-4,O). Let C1 denote the circle (x - 212 y2 = 9; let C2 denote the circle (x 2)' y2 = 9; let C3 denote the circle x2 y2 = 25. Given that

+ +

Pdx

+

+ Q d y = 11,

Pdx

+ Q d y = 9,

c2

CI

+

P d x + Q d y = 13, c3

-

f.r

..

find fpdx

+Q ~ Y .

c 4

+

where Cq is the circle x2 y2 = 1. [Hint: Use the result of Problem 8(a).] 10. Let F(x, y) = x 2 - y2.Evaluate a) \

:4

VF . dr on the c w e y = x3;

+

d s on the circle x 2 y2 = I, if n is the outer normal and = VF - n is the directional derivative of F in the direction of n (Section 2.14). 11. Let F(x, y) and G(x, y) be continuous and have continuous derivatives in a domain D. Let R be a closed region in D with directed boundary BR consisting of closed curves

b)

1

Chapter 5 Vector Integral Calculus C1,. . . , C,, as in Fig. 5.21. Let n be the unit outer normal of R and let aF/an, aG/an denote the directional derivatives of F and G in the direction of n: aF/an = V F . n, aG/an = V G . n. d s = SIRv 2 F dx d y . a) Show that lBR b) Show that lBR V F . d r = 0. IS,

C) Show that if F is harmonic in D, then JBR d) Show that

d s = 0.

+

[Hint: Use the identity div(f u) = f div u grad f . u of Section 3.6.1 12. Under the assumptions of Problem 11, prove the identities: a) lBR - ~ g ) d =s l S R ( ~ v-2G~ V ~ F ) ~ X ~ ~ [Hint: Use Problem 1l(d), applied to F and G and then with F and G interchanged]; b) if F and G are harmonic in R, then

(FE

Problems 1l(d) and 12(a) are known as identities of Green. Generalization to space and further applications are considered in Problem 4 following Section 5.11.

Remark

( A

r

t

PART 11. THREE-DIMENSIONAL THEORY AND APPLICATIONS 5.8 LINEINTEGRALS IN SPACE The basic definitions (Sections 5.2 and 5.3) of the line integrals

can be generalized almost without change to give corresponding line integrals along a curve C in xyz-space. Let C be given parametrically by equations:

and let C be directed by increasing r. We assume that +(t), $(t), ~ ( tare ) continuous and have continuous derivatives over the interval, so that C is "smooth." The interval h 5 t 5 k is now subdivided by the successive points to = h , t l , . . . , ti, . . . , t, = k. The value ti* is chosen between ti-, and ti, and one sets x,* = +(ti'), y,* = @(ti'), zf = o(t:). One denotes by h i t the difference ti - ti-, and by Ais the length of the corresponding piece of C. One defines

303

304

Advanced Calculus, Fifth Edition

i

A,,

Figure 5.24

Line integral in space.

These quantities are indicated in Fig. 5.24. One can now make the definition r

n

4

The integrals

br

are defined in similar fashion. Finally, J~(x.y,z)ds=

~ ~ ~i ( x% :,y:,z:)~,s. maxA,s+O

C

(5.62)

I=[

These definitions can again be shown to be meaningful if the functions X, Y, Z are continuous on C. The integrals can be reduced to ordinary definite integrals as in the plane:

One can also extend the definitions and results to arbitrary "piecewise smooth" curves C, that is, curves made up of a finite number of smooth pieces. The x - , y-, and z-integrals will usually be considered together:

/

ri.

X d x + Ydy + Z d z .

C

and this combination is related to an s-integral by the equation: /xdx+~dy+2dz= C

J

(Xcoso+Ycos,9+Zcosy)ds,

C

where

T = cos cri

+ cos Bj + cos y k

.

I ,

r

.:.

(5.64)

305

Chapter 5 Vector Integral Calculus is the unit tangent vector in the direction of increasing s. Equation (5.64)has a vector interpretation:

where the vector line integral on the left is the limit of a sum

and U T = u . T is the tangential component of the vector

u = Xi + Yj

+ Zk.

ir

3

...

F

The integrals in (5.64) and (5.65) all measure the work done by the force u on the given path. The interpretation i v , ds, in terms of a normal component, does not generalize to line integrals in space but does to surface integrals, to be defined later.

5.9 SURFACES IN SPACEH ORIENTABILITY

t

Let a surface S be given in space. The surface may be represented in the form:

z

= f ( x , y),

(x, y )

in region R,,.

(5.6G

in the form

11

or even in a parametric form (Section 4.7): x = f (u, v ) ,

,

y = g(u, v ) ,

,

*

.

z = h(u, v ) , ( u , v ) in R,,.

(5.68)

It will be assumed that S is "smooth," that is, that in (5.66),for example, f ( x , y ) has continuous derivatives in a domain containing the region R,,. It will be assumed that R,, is bounded and closed. It is then possible to evaluate the area of S or of a part of S. If the surface is given in form (5.66), the area of S is given by the formula (Section 4.7):

where y is the angle between the upper normal to S and the z direction, as in Fig. 5.25. Thus 0 5 y < 5 and sec y > 0. We can regard formula (5.69)as stating that the surface area element da for a surface z = f ( x , y) is

306

Advanced Calculus, Fifth Edition

Figure 5.25

-r

Ji

Surface integ;al.

d

1

If the surface is given in form (5.68), the area is given by

where the subscripts denote partial derivatives (Section 4.7). It is assumed here that the correspondence between parameter points ( u , v) and surface points ( x , y, z) is one-to-one and that E G - F~ # 0. Corresponding to (5.71), one has the sulface area element

The preceding formulas can be extended to surfaces S that are piecewise smooth, that is, pieced together of a finite number of smooth parts. One can compute the total area by adding the areas of the parts. For surfaces in implicit form F ( x , y , z ) = 0, one can in general decompose the surface into smooth pieces in each of which a representation z = z(x, y) or x = x(y, z ) or y = y ( x , z ) is available, so that the area can be computed by several applications of the formula (5.69) (cf. Problem 9 following Section 4.7). In order to define a surface integral analogous to the line integral SX d x Y dy Z dz it is necessary to assume that the surface S has a "direction assigned." One does not ordinarily consider the possibility of assigning a direction to a surface, but something equivalent to this is not unfamiliar, namely, the choice of a positive direction for angles, as in the xy-plane. For a general smooth surface S in space a simple way of achieving this is by choosing a unit normal vector n at each point of S in such a manner that n varies continuously on S . If this field of normal vectors has been chosen we term the surface oriented. One can then define a positive direction for angles at each point of the surface, as suggested in Fig. 5.26(a). When this has been done, one can also assign positive directions to simple closed curves forming the boundary of S. For each boundary curve C , one can choose a tangent T in the direction chosen and an inner normal N, in a plane tangent to S, as in Fig. 5.26(a). The vectors T, N, n must then form a positive triple at each point of C.

+

+

307

Chapter 5 Vector Integral Calculus

(a)

(b)

Figure 5.26 (a) Oriented smooth surface. (b) Miibius strip, a nonorientable surface.

Figure 5.27

Oriented piecewise smooth surface (cylinder).

If the surface S is only piecewise smooth, as is, for example, the surface of a cylinder, one cannot choose a continuously varying normal vector for all of S. In this case we say that S has been oriented if an orientation has been chosen in each smooth piece of S, in such a manner that along each curve that is a common boundary of two pieces the positive direction relative to one piece is the opposite of the positive direction relative to the other piece. This is illustrated for a cylindrical surface in Fig. 5.27. If a piecewise smooth surface S can be so oriented, we term the surface S orientable. It should be remarked that not every surface is orientable. Figure 5.26(b) suggests a Mobius strip, which is nonorientable. For this surface S, one easily convinces one's self that the normal n cannot be chosen to vary continuously on S and that the surface has the peculiar property of having only "one side." One-sidedness and nonorientability go together for surfaces in space.

i

308

Advanced Calculus, Fifth Edition

4

For surfaces z = f (x, y) there is no difficulty in this regard, since one can always choose n as the upper normal or, alternatively, as the lower normal. For surfaces in parametric form there is also no difficulty. It follows from the discussion of Section 4.7 that the vector

PI x P2 = (x,i

+ y,j + z,k) x (x,i + y,j + z,k)

is a normal vector having magnitude d

m .One can hence use

throughout or the negative of this vector throughout, provided that E G - F~ # 0. For a surface defined by an implicit equation F(x, y, z) = 0, one can choose n as VFIIVFI (or its negative), provided that V F # 0.

Let S be a smooth surface as previously and let H(x, y, z) be a function defined and continuous on S. Then the surface integral of H over S is defined as follows:

The surface S is here assumed cut into n pieces as in Fig. 5.25, A, a denotes the area of the ith piece, and it is assumed that the ith piece shrinks to a point as n + oo in an appropriate manner. The subdivision and the limit process are best described in terms of a representation (5.66) or (5.68), in which case they are precisely the same as the procedures used for the double integral in Section 4.3. If the representation z = f (x, y) is used, the integral is reducible to a double integral in the xy-plane:

where y is the angle between the upper normal and the z axis. This follows from the area formula (5.69) or, more concisely, from the formula da = sec y dx dy. If the parametric representation (5.68) is used, the integral has the expression:

m du d v for the area This follows similarly from the expression da = d element. The existence of the surface integral (5.74) and the expressions (5.75) and (5.76) can be established when H is continuous. As was pointed out in Section 4.7, the notion of surface area is an awkward one. Once that has been settled and the formula

309

Chapter 5 Vector Integral'Calculus for the area element justified, the proof of (5.75) and (5.76) parallels that for line integrals. The integral jJ H d a is analogous to the line integral j" F ds. In order to obtain a surface integral analogous to the line integral [ X dx Y dy Z dz we consider an oriented smooth surface S, with continuous unit normal vector n:

+

+

.\

n = cos cri + cos pj + cos yk.

Let L ( x , y, z), M ( x , y , z ) , N ( x , y , z ) be functions defined and continuous on S. Then by definition

i

If these are added, we obtain the general surface integral

I

This can at once be written in terms of vectors:

/SL

+

I

+

thus the surface integral dy dz M dz dx N dx dy is the ind$al over the surface of the normal component of the vector v = Li Mj N k . For evaluation of the surface integral, one has the following formulas:

I. If S is given in the form with normal vector n, then

+

+

, --

,

,.

iL .!

(5.80) Rc,

+

with the sign when n is the upper normal and the - sign when n is the lower normal. is

I

310

Advanced Calculus, Fifth Edition

11. If S is given in parametric form x = f (u,v),

z

. y = g(u, v),

= h(u, v), (u,v) in R,,,

with normal vector n, then

//

+

+

~ d y d z ~ d z d x~ d x d y

%

Z

S

with the

+ or - sign accordingly as

The proof of (5.80)is given here, the proof of (5.81)being left as an exercise [Problem 8(c)]. For a surface z = f ( x , y) the methods of Section 2.13 show that the tangent plane at (XI , y 1, z I ) is

and hence that the unit normal vector is

+

with the or - sign according to whether n is upper or lower. The denominator here is sec y ' , where y' is the angle between the upper normal and k. Thus

n = f

-,i

az

az. - FJ +k

sec y'

S ' -7

R,,

since d o = sec y ' d x d y . Accordingly, (5.80)is proved.

1

311

Chapter 5 Vector Integral Calculus

(a)J N dx dv = IJN dr dy. S R

(b)

IN dr dy = - I ~ Ndx dv. S

R E

Figure 5.28 Double integral over R versus surface integral over S,S = R w ~ t horientation. b

*

When L = 0 and M = 0, (5.80) reduces to the formula

SIN

Thus the surface integral d x dy over S is the same as plus or minus the double integral of N over the projection of S on the xy-plane, with plus or minus sign according to whether the chosen normal on S is upper or lower. In particular, S can be chosen as a region R in the xy-plane, and N = N(x, y). Thus the surface integral SS, N(x, y) dx dy and the double integral SS, N(x, y) d x dy are not the same; they are equal when S has the orientation determined by k and are otherwise negatives of each other, as illustrated in Fig. 5.28. This situation indicates how one can enlarge the concept of double integral to permit a choice of direction of integration. If this is done, Green's theorem can be stated in the form

without a direction of integration indicated on C but with the understanding that the "directions" on C and S match, that is, that when C is positively directed the normal associated with S is k, but when C is negatively directed the normal associated with S is -k. These two cases are shown in Fig. 5.28. The surface integral S S L dy dz + . . . can also be defined in a manner paralleling the vector line integral S u . d r of (5.65). One introduces a differential area vector do by the equation d o =nda, so that d o = cosa d a i

and

+ cos

d a j + cosy d a k

I

.

r

Advanced Calculus, Fifth Edition

312

1 One can also define this as the limit of a sum, as for the line integral. For the surface z = f (x,y), one finds a2

I

1

a2

--i--j+k ax ay

4

This can also be written:

n

d u = n l s e c y I d x d y = -d x dy. In. kl From this and the corresponding expressions for surfaces x = g(y, z), y = h ( x , z), one obtains the formulas

For the surfaces in parametric form we have

or, more concisely, by (5.82), d u = f( P I x P2) du dv. The formal properties of line and surface integrals are analogous to those for line integrals in the plane (Section 5.3) and need no special discussion here. Furthermore, the definitions and properties of line and surface integrals carry over without change to piecewise smooth curves and surfaces, provided that the surfaces are orientable.

PROBLEMS 1. Evaluate the line integrals: a) ~ ~ b q ~ ) z d x + ~ d w ~ h+e r~e cdi szt h.e c u r v e x = c o s t , y = sint, z = t , C

I'

OIt52n;

b)

c<':)

x2 dx - x i dy

+ y2 dz on the straight line joining the two points;

d) j c u ~ d s ,where u = 2xy2zi+2x2yzj+x2y2kand C is thecirclex = cost, y = sin t , z = 2, directed by increasing t ; e) J, U T ds, where u = curl v, v = y2i + z2j + x2k, and C is the path x = 2t 1, y = t2, z = 1 + t3, 0 5 t 5 1, directed by increasing t. 2. If u = grad F in a domain D, then show that

+

(~2.4'2.~~) X I ~ Y I ~ ZUI T)

J(

d s = F(x2, y2, z2) - F(xl, y1, ZI), where the integral is along any path in D joining the two points; U T d s = 0 on any closed path in D. b) a)

3

$,

I

313

Chapter 5 Vector Integral Calculus

3. Let a wire be given as a curve C in space. Let its density (mass per unit length) be S = S(x, y, z ) , where ( x , y, z ) is a variable point in C . Justify the following formulas: a) length of wire = /, ds = L;

b) mass of wire = Jc S ds = M; C) center of mass of the wire is (X,j , i ) ,where

1

d) moment of inertia of the wire about the z axis is I, =

J

I

+

( x 2 y 2 )S d s .

C

4. Formulate and justify the formulas analogous to those of Problem 3 for the surface area, mass, center of mass, and moment of inertia of a thin curved sheet of metal forming a surface S in space. 5. Evaluate the following surface integrals: a) SSs x d y dz y dz dx z dx d y , where S is the triangle with vertices (1, 0, O), (0, 1, O), ( 0 , 0 , 1 ) and the normal points away from (O,0,0);

+

b)

+

//, dy dz + dz dx + dx d y , where S is the hemisphere z = d m ,x2 +

y2 ( 1 , and the normal is the upper normal; c) SSs(x cos a y cos z cos y ) d u for the surface of part (b);

+

+

+

d) SSs x2z d o , where S IS the cylindrical surface x2 y2 = 1 , 0 5 z 5 1 . 6. Evaluate the surface integrals of Problem 5, using the parametric representation: a)x=u+v, y=u-v, z=1-2u b) x = sin u cos v, y = sin u sin v, z = cos u C) same as (b) d) x = c o s u , y = s i n u , z = v 7. Evaluate the surface integrals: a) SS, w . n d u , if w = xy2zi - 2x3j yz2k, S is the surface z = 1 x2 - y2, x2 y2 5 1 , and n is upper; b) SSs w . n d u , if w = i 2j 3k, S is the surface x = eU cosv, y = eu sinv, z = cos v sin v, 0 5 u 5 1 , 0 5 v 5 n/2, and n is given by (5.82) with the + sign;

+

-

+

+ +

9d a if w = x

C) ISS

~ and~S and ~ n are z as in (a);

SSs 2 d a if w = n2 - y2 + z2 and S and n are as in (b); e) //,curlu . n d u if u = yzi - xzj + xzk, S is the triangle with vertices ( 1 , 2, 8),

i *

rl

d)

(3, 1, 9). (2, 1, 7) and n is upper. 8. a) Let a surface S: z = f ( x , y) be defined by an implicit equation F ( x , y , z) = 0. show^ that the surface integral H d a over S becomes

provided that

# 0.

314

I

Advanced Calculus, Fifth Edition

b) Prove that for the surface of part (a) with a = VF/(VFI,

,

.4

c) Prove (5.81). d) Prove that (5.81) reduces to (5.80) when x = u , y = v , z = f ( u , v). 9. Let S be an oriented surface in space that is planar; that is, S lies in a plane. With S one can associate the vector S, which has the direction of the normal chosen on S and has a length equal to the area of S. a) Show that if S1, S2, S3, S4 are the faces of a tetrahedron, oriented so that the normal is the exterior normal, then S1 S2 S3 S4 = 0. [Hint: Let S, = A,n, (Ai > 0) for i = 1, . . . , 4 and let S1 . . . -t S4 = b. Let pl be the foot of the altitude on face S1 and join pl to the vertices of S1 to form three triangles of areas A I 2 , . . . , A I 4 .Show that, for proper numbering, A l j = fA,n, .nl, with or - according as n, .nl > 0 or < 0, and A , , = 0 if nj . nl = O ( j = 2,3,4). Hence deduce that b . n, = 0 for j = 2 , 3 , 4 and thus b . b = 0.1 b) Show that the result of (a) extends to an arbitrary convex polyhedron with faces S1, . . . , S,, that is, that SI + s ~ + . . . + S n=0, when the orientation is that of the exterior normal. c) Using the result of (b), indicate a reasoning to justify the relation

+ + +

+

+

for any convex closed surface S (such as the surface of a sphere or ellipsoid), provided that v is a constant vector. d) Apply the result of (b) to a triangular prism whose edges represent the vectors a, b, a b, c to prove the distributive law (Equation (1.19) cx(a+b)=cxa+cxb for the vector product. This is the method used by Gibbs (cf. the book by Gibbs listed at the end of this chapter).

+

It was pointed out in Section 5.5 that Green's theorem can be written in the form

T h e following generalization thus appears natural:

11

u, da =

S

///

div v dx dy d l ,

R

where S is a surface forming the complete boundary of a bounded closed region R in space and n is the outer normal of S , that is, the one pointing away from R.

il

Chapter 5 Vector Integral Calculus The theorem, known as the Divergence theorem or Gauss's theorem, plays a role like that of Green's theorem for line integrals, in showing that certain surface integrals are 0 and that others are "independent of path." The theorem has a basic physical significance, which will be considered later. If one considers a continuous flow of matter in space, the left-hand side can be interpreted as the flux across the boundary S, that is, the total mass leaving R per unit time; the right-hand side measures the rate at which the density is decreasing throughout R, that is, the total loss of mass per unit time. Under the assumptions made, the only way mass can be lost is by crossing the boundary S. For this reason the flux equals the total divergence.

+ +

DIVERGENCE THEOREM (Gauss's Theorem) Let v = Li Mj Nk be a vector field in a domain D of space; let L , M, N be continuous and have continuous derivatives in D. Let S be a piecewise smooth surface in D that forms the complete boundary of a bounded closed region R in D. Let n be the outer normal of S with respect to R. Then

11

v,, d o =

Ifl

div v dx dy dz;

that is,

Proof. It will be proved that

jj

Ndxdy

J J J

dZ

the proofs of the two equations

being exactly the same. It should first be remarked that the normal n, defined as the outer normal of S with respect to R, necessarily varies continuously on each smooth part of S, so that the surface integral in (5.85) is well defined. The orientation defined by n on each part of S in fact determines an orientation of all of S, so that S must be orientable. We assume now that R is representable in the form: f i b 7

Y ) i Z i f2(x3Y ) ,

( x , Y ) in RxY,

(5.89)

316

Advanced Calculus, Fifth Edition

Figure 5.29

Proof of Divergence theorem.

where R,, is a bounded closed region in the xy-plane (cf. Fig. 5.29) bounded by a simple closed curve C. The surface S is then composed of three parts: Si: z = f ~ ( xy), ,

(x, y) in R,,, ,

I

S2: z = f 2 ( ~Y), , (x, y) in R,,, S3: f i ( ~ ,y) i z i f 2 ( ~Y), , (x, y) on C.

S2forms the "top" of S, S1forms the "bottom," and S3forms the "sides" (the portion S3 may degenerate into a curve, as for a sphere). Now by definition (5.77),

where y = c ( n , k). Along S3, y = n/2, SO that cos y = 0;along S2, y = y' where y' is the angle between the upper normal and k; along S1, y = n - y'. Since d a = sec y ' d x d y on S1 and SZ,one has

+

/ / ~ d ~ =d/ /~ ~ d x d ~ S m

i

f . .

SI

z

~

s 2

=-//Ncosy'secy'dxdy+ R,,

where

//

d

+

//

1

h

1

~ ~ d dx d y ~

11

Ncosy'secy'dxdy.

R,,

= f ~ ( x y) , in the first integral and z = f2(x, y) in the second:

4

Chapter 5 Vector Integral Calculus Now the triple integral on the right-hand side of (5.86) can be evaluated as follows:

I

This is the same as (5.90); hence

//

N dx dy =

/// a

dx dy d t . ~ s -

.

TC~T

R

S

This is the result desired for the particular R assumed. For any region R that can : be cut up into a finite number of pieces of this type by means of piecewise smooth auxiliary surfaces, the theorem then follows by adding the result for each part separately (cf. the proof of Green's theorem in Section 5.5). The surface integrals over the auxiliary surfaces cancel in pairs, and the sum of the surface integrals is precisely that over the complete boundary S of R; the volume integrals over the pieces add up to that over R. The result can finally be extended to the most general R envisaged in the theorem by a limit process [see the book by Kellogg, listed at the end of the chapter]. Once (5.86) has been established, (5.87) and (5.88) then follow by merely relabeling the variables. Adding these three equations, one obtains Eq. (5.85). As a first application of the Divergence theorem, a new interpretation of the divergence of a vector field v will be given. Let Sr be a sphere with center (x, , yl , z l ) and radius r. Let Rr denote S, plus interior. We recall that the Mean Value theorem $0 for integrals asserts that

where (x*, y*, z*) is a point of R and V is the volume of R, provided that f is continuous in R (cf. the formulation for double integrals in Section 4.3). If this is applied to the integral

/fl

div v dx d y dz.

Rr

one concludes that

J J J Rr

I1

318

Advanced Calculus, Fifth Edition

,

for some (x*, y*, z*)in R,, Vr denoting the volume (:7rr3)of R,. If the Divergence theorem is applied, one concludes that

If now we let r approach 0, the point (x*, y*, z*) must approach the center (x, , yl , zl). Thus

The surface integral S f v n d a can be interpreted as theJlux of the vector field v across S,. Hence the divergence of a vectorjield at a point equals the limiting value o f t h e m across a sphere about the point divided by the volume of the sphere, as the radius of the sphere approaches 0. The spherical surface and solid used here can be replaced by a more general one, provided that the "shrinking to zero" takes place in a suitable manner. Thus in concise form, divergence equals flux per unit volume. This result has fundamental significance, for it assigns a meaning to the divergence that is independent of any coordinate system. It also provides considerable insight into the meaning of positive, negative, and zero divergence in a physical problem. One can use (5.91) as the definition of divergence, and this is often done; one must then verify, with the aid of the Divergence theorem in the form (5.85),that one obtains the usual formula

in each coordinate system. One can apply (5.91) to the case in which v = u, where u is the velocity vector of a fluid motion. The expression u, d a can be interpreted as the volume filled up per unit time by the fluid crossing the surface element do in the direction of n (see Fig. 5.30). The flux integral [Sun d o over the surface S, then measures the total rate of filling volume (outgoing minus incoming volume per unit time). By (5.91), div u then measures the limiting value of this rate, per unit volume, at a point (x,, yl, zl). If the fluid is incompressible, the outgoing volume precisely equals the incoming volume, so that the flux across S, is always 0. By (5.91) this implies that div u = 0; conversely, by (5.84), if div u = 0 in D, then the flow is incompressible. Another interpretation of div u is given in Problem 7 following Section 5.15. One can also let v = pu, where p is the density and u is the velocity of the fluid motion; the flux integral

then measures the rate at which mass is leaving the solid R via the surface S, for u, d a measures the volume of fluid crossing a surface element per unit of time, and pun d a measures its mass. The divergence div v measures this rate of loss of

Chapter 5 Vector Integral Calculus n

Figure 5.30

Flux = ffu,, d o .

mass per unit volume, by (5.91). But this is precisely the rate at which the density is decreasing at the point ( x l, yl , zl). Hence

or

a~

at

+ div (pu)= 0.

(5.92)

This is the continuity equation of hydrodynamics. It expresses the conservation o f mass. Another derivation is given in Problem 9 following Section 5.15.

PROBLEMS 1. Evaluate by the divergence theorem: a) ISs x d y d z + y d z d x + z d x d y , where S is the sphere x 2 y2 z2 = 1 and n is the outer normal; b) Sjs v,, d o , where v = x2i y2j z 2 k , n is the outer normal and S is the surface of t h e c u b e O s x 5 l , O s y 5 1 , O i z s 1; C) e Ycos z d y d z ex sinz d z d x ex cosy d x d y , with S and n as in (a);

+ +

+

+

ISS + + d) ISs VF . n d o if F = x 2 + y2 + z2, n is the exterior normal, and S bounds a solid e)

region R; JJs V F . n d a if F = 2x2 - y2 - z2, with n and S as in (d);

f ) f f, VF . n d a if F = [ ( x - 212 + y2 + z ~ ] - ' /and ~ S and n are as in (a). 2. Let S be the boundary surface of a region R in space and let n be its outer normal. Prove the formulas:

a) V = S J s x d y d z = j j s y d z d x = j j s z d x d y

= ~ ~ ~ ~ x d yvd zdd xz ++z d x d y , where V is the volume of R;

320

Advanced Calculus, Fifth Edition

11,

+

{

3

-j

+

x2 dy dz 2xy dz d x 2x2 d x dy = 6V.f. where ( f ,j , 2) is the centroid of R; c) jfs curl v . n d a = 0, where v is an arbitrary vector field. 3. Deduce the results of Problem 9(a), (b), (c) following Section 5.10 by proving (c) first, using the incompressibility of the flow of constant velocity v. 4. Let S be the boundary surface of a region R, with outer normal n, as in the Divergence theorem. Let f (x, y, z) and g(x, y, z) be functions defined and continuous, with continuous first and second derivatives, in a domain D containing R. Prove the following relations:

b)

a)

j

14

ffsfagian d o = jfif v 2 g d~ dy dz + j & ( v f .

- +

[Hint: use the identity V . (f u) = Vf u b) if g is harmonic in D, then

v g ) d dy ~ dz; f ( V . u).]

[Hint: Put f = 1 in (a).] c) i f f is harmonic in D, then

d) if f is harmonic in D and f = 0 on S, then f E 0 in R [cf. the last paragraph before the remarks at the end of Section 4.31; e) i f f and g are harmonic in D and f = g on S, then f = g in R; [Hint: Use (d).] f) i f f is harmonic in D and af/an = 0 on S, then f is constant in R; g) if f and g are harmonic in D and aflan = aglan on S, then f = g const in R; h) if f and g are harmonic in R, and

1

+

then f

E

g in R;

i) if f and g both satisfy the same Poisson equation in R ,

v2f= -4nh,

v~~= -4nh,

h =h(x, y , ~ ) ,

and f = g on S, then f ~ g i n R ;

j)

jh (f 8 - i g ) d o = f f j R ( f v 2- g v 2 f) d x dy dz;

[Hint: Use (a).] k) if f and g are harmonic in R, then

4

Chapter 5 Vector Integral Calculus

321

1) i f f and g satisfy the equations:

v2f= h f ,

V 2g = h g ,

h=h(x,y,z),

in R, then

Remark

Parts (a) and ( j ) are known as Green'sjrst and second identities, respectively.

5. Let S and R be as in Problem 4. Prove, under appropriate continuity assumptions: a)

JS, f n . i d o =luR g d

b)

SJSfn d o = JURV f d V.

L

~ . [Hint: Apply the Divergence theorem.]

[Hint: These are integrals of vectors as in Section 4.5. Use (a) to show that the x-components of both sides are equal and, similarly, that the y- and z-components are equal.] C) JSSv x i . n d u = l & c u r l v . i d V . [Hint: Apply the Divergence theorem and then evaluate div (v x i) by (3.35).] d) l j s n x v d o = jjJRcurl vdV. [Hint: These are vector integrals. Use (c) to show that the x-components of both sides are equal and, similarly, that the y- and z-components are equal.]

It was seen in Section 5.5 that Green's theorem can be written in the form

This suggests that for any simple closed plane curve C in space (Fig. 5.31),

/

ur ds =

//

curl. u do,

(5.93)

where n is normal to the plane in which C lies, S is the planar surface bounded by C, and the direction of C is positive in terms of the orientation of S determined by n.

Figure 5.31

Case of Stokes's theorem.

322

Advanced Calculus, Fifth Edition z

Figure 5.32

Another case of Stokes's theorem.

Figure 5.33

Proof of (5.94).

It was pointed out in Section 5.9 that the choice of a continuously varying normal vector on a surface determines a positive sense for each simple closed curve C, when regarded as bounding a portion of the surface. Equation (5.93) can be generalized further, namely, by replacing S by an arbitrary smooth orientable surface whose boundary is a simple closed curve C, not necessarily a plane curve. Again the choice of a continuous normal determines the direction on C, as in Fig. 5.32. It should be stressed that this relation between n and the direction of C, as defined in Section 5.9, depends on the notion of positive triple of vectors, hence on a particular orientation of space (Section 1.1). If the orientation were reversed, the given normal n would correspond to the opposite direction on C. Equation (5.93) is Stokes's theorem, which will be shown to be correct under very general assumptions. An additional reason for expecting such a relation as this is the fact that if two surfaces S1 and S2 have the common boundary C and , no other points in common, as in Fig. 5.33, then

provided that S1 and S2 are oriented so as to determine the same direction on C.

I

1

3

Chapter 5 Vector Integral Calculus For

where S is the oriented surface formed of S1 and of S2 with its normal reversed. S then bounds a certain solid region R and by the Divergence theorem

/I

curl# u do =

+

10

div curl u d x dy dz = 0.

(The + or - depends on whether n is outer or inner normal.) Thus the surface integral JJ curl, u d A has the same value for all surfaces with boundary C, and it is natural to expect the surface integrals to be expressible in terms of a line integral of u on C.

THEOREM OF STOKES Let S be a piecewise smooth oriented surface in space, whose boundary C is a piecewise smooth simple closed curve, directed in accordance with the given orientation in S . Let u = Li M j Nk be a vector field, with continuous and differentiable components, in a domain D of space including S. Then

+ +

/

C

u r ds = \/(curl

u . n) do,

S

(5.95)

-

< < Z

,

where n is the chosen unit normal vector on S ; that is,

Proof. Just as with the Divergence theorem, it is sufficient to prove three separate equations:

One can further restrict attention to the case in which S is representable in the form z=f(x,y>,

(x,y)inRxy,

as shown in Fig. 5.34; as in the proof of the Divergence theorem, the general case is handled by decomposing into a finite number of such pieces, when possible, and by a limit process. If S has the form z = f (x, y), the curve C has as projection on the xy-plane a curve C,, as shown in Fig. 5.34; as (x, y, z) goes around C once in the given direction, its projection (x, y , 0) goes once around C,, in a corresponding direction.

324

Advanced Calculus, Fifth Edition

Figure 5.34

Proof of Stokes's theorem. 1'4

If the normal n chosen on S is the upper normal, the direction on Cxyis the positive direction, and by Green's theorem, / L ( ~ . Y . Z )= ~ /~L I X , Y , f(x, y)]dx = C

C x ~

-11(5 "%) +

R ~ Y

dx dy.

az ay

Under the same assumption about the normal,

by (5.80). It at once follows that

If the direction of n is reversed, both sides change sign, so that (5.97) holds in general. By the reasoning described above, one then finds that (5.97) holds for a general orientable S. In the same way, equations analogous to (5.97) are established for M and N for a general S. Upon adding the equations for L, M. and N, one obtains Stokes's theorem in full generality. Just as the Divergence theorem gives a new interpretation for the divergence of a vector, so does the Stokes's theorem give a new interpretation for the curl of a vector. To obtain this, we take S, to be a circular disk in space of radius r and center (x,, y ~2,1 ) bounded by the circle C , (see Fig 5.35). By the Stokes's theorem and Mean Value theorem for integrals,

4

Chapter 5 Vector Integral Calculus

Figure 5.35

Meaning of curl.

where A, is the area (nr2)of S, and (x*, y*, z*) is a suitably chosen point of S,. One can now write curI,u(x*, y*, z * ) = c,

In the case of a fluid motion with velocity u the integral jcru~ ds is termed the circulation around C,; it measures the extent to which the corresponding fluid motion is a rotation around the circle C , in the given direction. If r is now allowed to approach 0, we find

that is, the component of curl u at ( x , , y , , z l ) in the direction of n is the limiting ratio of circulation to area for a circle about (xi, y l , z l ) with n as normal. Briefly, curl equals circulation per unit area. In the limit process the circular disk can be replaced by a more general surface with normal n, provided that the shrinking to zero is properly carried out. If n is taken as i, j, and k successively, one obtains the three components of curl u along the axes. Since (5.98) has a significance independent of the coordinate system chosen, this equation proves that the curl of a vector field has a meaning independent of the particular (right-handed) coordinate system chosen in space. One could in fact use (5.98) to define the curl. [If the orientation of space is reversed, the direction of the curl will also be reversed (cf. also Section 3.8).] I"

5.13 INTEGRALS INDEPENDENT OF PATHw IRROTATIONAL AND SOLENOIDAL FIELDS

- i? ;

Since the generalization of Green's theorem to space takes two different forms, the Divergence theorem and Stokes's theorem, one can generalize the discussions of Sections 5.6 and 5.7 in two directions: to surface integrals and to line integrals. In the case of line integrals the results for two dimensions carry over with minor . modifications. The surface integrals require a somewhat different treatment. Line integrals independent of path in space are defined just as in the plane. The following theorems then hold.

325

B 326

Advanced Calculus, Fifth Edition

+ +

THEOREM I Let u = Xi Y j Zk be a vector field with continuous components in a domain D of space. The line integral

is independent of path in D if and only if there is a function F ( x , y, z), defined in D such that

throughout D. In other words the line integral is independent of path if and only if u is a gradient vector:

u = grad F. The proof for two dimensions in Section 5.6 can be repeated without essential change. When the integral is independent of path, X d x Y d y Z d z = dF for some F , and

+

+

4

as in the plane.

THEOREM I1

Let X, Y, Z be continuous in a domain D of space. The line integral ,

Jxdx+ydy+zdz is independent of path in D if and only if

on every simple closed curve C in D. This is proved as in the plane.

+ +

THEOREM I11 Let u = Xi Yj Zk be a vector field in a domain D of space; let X, Y, and Z have continuous partial derivatives in D. If

is independent of path in D, then curl u

= 0 in D; that is,

Chapter 5 Vector Integral Calculus

+

+

Conversely, if D is simply connected and (5.99) holds, then S X dx Y dy Z dz is independent of the path in D; that is, if D is simply connected and curl u = 0 in D, then

327

I

r7

u = grad F for some F. A domain D of space is called simply connected if every simple closed curve in D forms the boundary of a smooth orientable surface in D. Thus the interior of a sphere is simply connected whereas the interior of a torus is not. The domain between two concentric spheres is simply connected, as is the interior of a sphere with a finite number of points removed. (This definition, which is adequate for practical applications, differs from the standard one of advanced mathematics.) In the first part of the theorem we assume that SuT.ds is independent of path. Hence by Theorem I, u = grad F. Accordingly, curl u = curl grad F

=0

by the identity of Section 3.6. In the second part of the theorem we assume that D is simply connected and curl u = 0. To show independence of path, it is sufficient, by Theorem 11, to show that J, ur ds = 0 on each simple closed curve C in D. By assumption, C forms the boundary of a piecewise smooth oriented surface S in D. Stokes's theorem is applicable, and one finds

for proper direction on C and normal n on S. We remark that the Stokes's theorem can be extended to an arbitrary oriented surface S whose boundary is formed of distinct simple closed curves C , , . . . , C,. If Bs denotes this boundary, with proper directions, one has

The proof is like that of Section 5.7. In particular, if curl u E 0 in D,

ls

uT

d s = 0.

This can be applied to evaluate line integrals in "multiply connected domains, as in Section 5.7.

R

328

Advanced Calculus, Fifth Edition

f

A vector field u (whose components have continuous derivatives) such that curl u

.I:,

=0

in a domain D,is called irrotational in D.By virtue of the above theorems, irrotationality in a simply connected domain is equivalent to each of the properties:

I

U T ds

= 0 for every simple closed curve in

D;

;.

c

S

4 3ii;

UT

ds is independent of the path in D;

A 3

.T 3

u = grad F in D.

A theory similar to the preceding holds for surface integrals. Rather than give a full discussion here, we confine our attention to the counterpart of the last statement in Theorem 111; for more details we refer the reader to the books of Brand and Kellogg listed at the end of the chapter. I

+ +

THEOREM IV Let u = Li Mj Nk be a vector field whose components have continuous partial derivatives in a spherical domain D.If div u = 0 in D:

then a vector field v = Xi

+ Y j + Zk in D can be found such that curl v

= u in D ;

that is,

Remark.

The theorem provides a converse to the theorem of Section 3.6: div curl u

= 0,

for it asserts that if div u = 0, then u = curl v for some v. However, while div u may be 0 in an arbitrary domain D l , the theorem provides v whose curl is u only , is, no one v serves for all of Dl. in each spherical domain D contained in D l that Actually, the proof to follow gives one v for all of D lwhen D lis the interior of a cube or of an ellipsoid or of any "convex" surface. The existence of v for all of D lfor more general cases can be established (see, for example, pages 203 ff. in the book by Lamb listed at the end of the chapter). Vector fields u satisfying the condition div u = 0 are termed solenoidal. The theorem amounts to the assertion that solenoidal fields are (in suitable domains) fields of the form curl v, provided that the components of u have continuous partial derivatives. The field v is not unique (Problem 5). '

Chapter 5 Vector Integral Calculus

Proof of the theorem. Let the spherical domain D have center at Po;for simplicity we assume Po to be the origin (0, 0, 0). If P l ( x l ,y l , z l ) is an arbitrary point of D, we set

where, on the right-hand side, x , y, and z are the following functions o f t :

As t varies from 0 to 1, the point ( x ,y , z ) varies from Po to P1 on the line segment Po P I ; hence ( x , y , z ) remains in D. We have now by Leibnitz's Rule and the chain rule (Sections 4.9 and 2.8):

and, similarly,

Accordingly,

Since div (Li

Now

+ M j + N k ) = 0, this can be written as follows:

329

330

Advanced Calculus, Fifth Edition 'i Accordingly,

This gives the first of (5.100). The other two equations are proved in the same way. The solution

v can be expressed in the compact form:

vix, y , z) =

I'

tu(xt, yt, z t ) x (xi

+ yj + zk)dt.

(5.103)

If the vector field u is homogeneous of degree n, that is,

u(.rt, y t , z t ) = t"u(x,y, 2) (Problem 11 following Section 2.8), the formula can b e simplified further: 1

tn''u(x, y, Z)x (xi

+ yj + zk) dr

1 -( u x ~ ) ,r = x i + y j + z k .

n+2 For more on this topic, see pages 487-489 of vol. 58 ( 1 95 1 ) and pages 409-442 of vol. 109 (2002) of the American Mathematical Monthly.

PROBLEMS

.

, *%a

a

l'"-

1. Evaluate by Stokes's theorem: a) Jc U T d.5, where C is the circle x 2 positive x, and u 1s the vcctor -3yi

+

+

+ y2 = 1, z = 2, directed so that y increases for + 3xj + k ;

b ) J, 2xv2zdx 2x2yzd y ( x 2 y 2 - 2 z ) d z around the curve x = cos t . v = sin t , z = slnt, 0 5 t 5 2n,directed w ~ t hincreasing r.

2. By showing that the integrand is an exact differential, evaluate a)

&':':$

y z dx

b)

+ xz d y + x y d z on any path;

sin yr dx Z

= t.

+ xz cos y z d y + x y cos yz d r on the helix x = cost.

1 = sin t ,

3. Let C be a simple closed plane curve in space. Let n = ai + bj + ck be a unit vector normal to the planc of C and let the direction on C match that of n. Prove that 2

/ ( b z - cy) d.x + ( C X - UZ.) d y

+ (oy - hx)d z

C

equals the plane area enclosed by C. What does the integral reduce to when C is in the xy-plane?

4. Let u = -i

x-

-V

+ v-

+A

XL7+

ing the circle ( x - 2)-

j

+ zk and let D be the interior of the torus obtaincd by rotat-

+ z2 = I , y = 0 about the z-axis. Show that curl u = 0 in D but ?)-

I

331

Chapter 5 Vector Integral Calculus

lc

+

U T d s is not zero when C is 'the circle x 2 y2 = 4, z = 0.Determine the possible U T ds on a path in D. values of the integral

):2$

5. a) Show that if v is one solution of the equation curl v = u for given u in a simply connected domain D, then all solutions are given by v + grad f , where f is an arbitrary differentiable scalar in D. b) Find all vectors v such that curl v = u if 6. Show that iff and g are scalars having continuous second partial derivatives in a domain D, then u=Vf x v g is solenoidal in D. (It can be shown that every solenoidal vector has such a reprk%entation, at least in a suitably restricted domain.) 7. Show that if llsundo = 0 for every oriented spherical surface S in a domain D and the components of u have continuous derivatives in D, then u is solenoidal in D. Does the converse hold? , 8. Let C and S be as in Stokes's theorem. Prove, under appropriate assumptions: a) l c f T . i d s = S S s n x Vf . i d u ; [Hint: Apply Stokes's theorem, taking u = f i. Evaluate curl u by (3.28).] b) lcf T d s = J j s n x Vf d u . [Hint: These are vector integrals, as in Section 4.5. Show by (a) that the x-components of both sides are equal and, similarly, that the y- and z-components are equal.] 9. The operator v x V is defined formally as (vxi + vyj vZk)x (Vxi + V y j Vzk) = (vyV, - uzVy)i . . . Show that, formally: a) ( V XV ) . u = v . V x u = v . c u r l u ; b) (Vx V ) x u = V,(V. U) - ( V . U)V,where V,(v. u) indicates that v is treated as vyVuy v,VuZ. constant: V,(v . u) = v,Vu, t 10. Let C and S be as in Stokes's theorem. Show with the aid of Problem 9:

+

+

+

+

+

The formula for change of variables in a double integral:

//

F ( x , y ) d x dy =

Rr,

F[f (u, v), g(u, v)l

du d v ,

(5.104)

R" ,>

#

is given in Section 4.6.In this section we shall give a proof of this formula under appropriate assumptions. We shall also indicate how widely the formula is applicable and shall explain the more general formula:

S

//

F ( x , y ) d x dy =

FIf (u. v), g(u, v)l-

a(x' du d v , a(u, V)

(5.105)

332

Advanced Calculus. Fifth Edition where S is the "degree" of the mapping of the boundary of R,, into the boundary of THEOREM I

4

The formula

is valid under the following assumptions:

a) R,, and R,, are bounded closed regions in the xy- and uv-planes bounded by piecewise smooth simple closed curves C x yand C,,, respectively. b) The function F(x, y) is defined and has continuous first derivatives in a circular domain D,, containing R,,. The functions x = f (u, v ) , y = g(u, v) are defined and have continuous second derivatives in a domain D,, including R,,; when (u, v) is in D,,, the point ( x , y) is in D,,. d) When (u, v) is on C,,, the corresponding point (x, y ) , x = f (u, v), y = g(u, v), is on C,,; as (u, v) traces C,,, once in the positive direction, (x, y) traces C x y once in the positive direction [corresponding to the sign in (5.106)]or negative direction [corresponding to the - sign in (5.106)]. C)

-3

+

The situation is illustrated in Fig. 5.36. It is to be stressed that the mapping from the uv-plane to the xy-plane is not assumed to be one-to-one except on the boundary, that when (u, v) is in R,, the corresponding point (x, y) need not be in Rxy,and that no assumptions are made about the sign of the Jacobian a(x, y)/a(u, v). To prove the theorem, we first remark that one can represent F(x, y) as follows:

in D x y ,where Q has continuous first derivatives. One need only take rx

I Figure 5.36

I

.,

Transformation of double integrals.

4

B

333

Chapter 5 Vector Integral Calculus

I

Green's theorem now gives

1

The line integral J Q d y can at once be written as a line integral in the uv-plane:

The differentials here should all be thought of as expressed in terms of a parameter t : dy = ( d y / d t ) d t ,du = ( d u / d t ) d t , d v = ( d v / d t ) d t .As t goes from h to k, C,, is to be traced just once in the positive direction, so that C,, is traced just once in the positive or negative direction. The f sign corresponds to these two cases. The line integral on the right is of form

where PI and Ql are defined in D,, and have continuous derivatives there. Hence Green's theorem is applicable:

since aQ/ax = F ( x , y). One thus concludes

JJ ~ ( xY ),d x d y = fJJ ~ R,,

as was to be shown.

R,,,.

a ( x 9 d u dv, [( u ,fv ) , g(u, U)I---. a(u9 v)

t

334

Advanced Calculus, Fifth Edition The restrictions made in the statement of the theorem can be relaxed considerably. First of all, it is not necessary to assume that, as (u, v) makes a circuit of C,,, the point ( x , y) moves steadily around C,,; the point ( x , y) can move ahead, then move backwards, then ahead again, and so on, as long as one complete circuit is achieved. One can make clear what is allowed and, in fact, get even greater generality as follows: Let 0 be a point inside C,, and let 0 be a polar coordinate angle measured relative to 0 , as in Fig. 5.36. As the point (u, v) goes around C,, in the positive direction, the angle 0 for the corresponding point ( x , y) can be chosen to vary continuously. When ( u , v) has made a circuit of C,,, 8 will have increased by a certain multiple 6 of 2n. This integer 6 is known as the degree of the mapping of C,, into C,,. With the theorem as stated previously, 6 = f1; for a general 6, the theorem must be restated:

THEOREM I1

The formula

is valid under the assumptions (a), (b), (c), of Theorem I and the following assumption: d') When (u, v) is on C,,, the corresponding point ( x , y ) is on C,,; the degree of the mapping of C,, into C,, is 6. The extension of the previous proof to this case causes no great difficulty. One need only verify that

this relation follows from the fact that, as (u, v) makes one circuit of C,, in the positive direction, the point (x, y) effectively traces the curve C,, 6 times. If S is negative, this means that ( x , y) traces C,, 161 times in the negative direction. It can happen that 6 = 0, in which case both sides of (5.107) are 0. It is of interest to note that a careful treatment of the transformation from rectangular to polar coordinates requires a formula as general as (5.107) (see Problem 4). One can also consider the case of a multiply connected region R,,, bounded by curves C&), C:;), . . . , C!&). If the boundary of R,, is a similar set of curves el:),c::)), . . . , c:",),then the formula (5.107) continues to hold, provided that each C;; is mapped on the corresponding C t i with the same degree 6. In particular, the points on these boundaries can be in one-to-one correspondence, with 6 always equal to 1; one then obtains formula (5.106) again, with the sign. The requirement that F ( x , y) be defined in a circular domain D,, was needed to show that a Q could be found such that a Q / a x = F. A more general domain would still permit this, but it is actually unnecessary to make any restriction whatsoever on the nature of D,,. The preceding proof clearly used the differentiability conditions on F, f , and g in an unavoidable manner. An entirely different method of proof can be devised that

+

4 3

335

Chapter 5 Vector Integral Calculus requires merely that F be continuous and that f (u, v) and g(u, v) have continuous first derivatives. Thus (5.107) remains valid if assumptions (b) and (c) are replaced by the following:

.,

-

'

*

b') The function F ( x , y) is defined and continuous in a domain D,, containing RX, .

c') The functions x = f (u, v), y = g(u, v ) are defined and have continuous derivatives in a domain D,, including R,,; when (u, v) is in D,,, (x, y ) is in D,,. For proofs the reader is referred to the articles "The Transformation of Double Integrals," by R. G. Helsel and T. Radb, in Transactions of the American Mathematical Society, Vol. 54 (1943), pages 83-102 and "Change of Variables in Multiple Integrals" by Peter D. Lax, in American Mathematical Monthly, Vol. 116 (1999), pages 497-50 1. When F(x, y) is chosen identically equal to 1 in (5.106), the left-hand side gives : the area A of R,,. Thus

~ = ~ / / m a(x' Y) d u dv.

If the Jacobian J = a(x, y)/a(u, v) is always positive or zero, the integral on the right is positive; hence only the sign can hold. We therefore conclude, on the basis of (a), (b), (c), (dl:

+

Ifthe Jacobian J is alwayspositive, then, as (u, v) traces C,, once in the positive direction, the point (x, y ) traces C,, once in the positive direction.

A similar result holds when J is negative, the positive direction on C,, being replaced by the negative direction. Combining the two cases, one obtains the following theorem:

THEOREM III The formula

holds under the assumptions (a), (b), (c), and (d), and e) The Jacobian a(x, y)/a(u, v) does not change sign in R,,.

It can be shown that if J # 0, the assumptions made for Theorem I11 imply that the mapping from the uv-plane to the xy-plane is one-to-one. Theorem I11 is thus essentially the standard form of the transformation theorem, as given in the books of Courant, Goursat, and Franklin listed at the end of the chapter. As was stated in Section 4.6, a formula analogous to (5.106) holds in three (or more) dimensions. The method of proof used here generalizes in a natural manner, the line integrals being replaced by surface integrals and Green's theorem by the

336

Advanced Calculus, Fifth Edition Divergence theorem. One can also define a degree 6 for the correspondence of boundaries, and a formula analogous to (5.107) then holds without any assumption about one-to-one correspondence on boundaries: s///F(x,

y , z ) d x d y dz = J/JFIX(U,

R,,:

U, w ) ,

...I a(x' y ' z )

acu, V , W )

d u d v d w . (5.109)

R,,,,I,

The degree 6 measures the effective number of times the bounding surface S,,; is traced by (x, y, z ) as the point (u, v , w) traces the bounding surface S,,, of R,,,,,. The surfaces S,,, and S,,, are both considered oriented, the normal being the outer normal for both; in measuring S, one counts negatively the parts of S,,, that are mapped into S,,, with reversal of orientation. Thus for the mapping

+ +

+ +

of the sphere u2 v2 w2 = 1 onto the sphere x2 y2 z2 = I , one has 6 = -1. Formula (5.109) can also be extended to the case in which R,,, and R,,; are each bounded by several surfaces, as in two dimensions. In the case in which R,,, is simply connected and is bounded by a single surface S,,, (which has then the structure of a sphere), the degree S can be computed, by analogy with the planar case, by reference to a "solid angle" (see Problem 6 ) .

PROBLEMS 1. Transform the integrals, using the substitution given: a ) J o' J 6v ( x2 + y 2 ) d ~ d y , u = y , v = x ;

3

b) J I R , ,( X - y ) d x d y , where R,?,is the region^*+^^ 5 1, andx = u + ( 1 - u2 - v 2 ) , y = v + (1 - u2 - v 2 ) ;(Hint: Use as R,, the region u2 v 2 ( 1.) c) JJR,, x y dx d y , where R,, is the region x2 y2 5 1 and x = u2 - v2, y = 2uv. [Hint: Choose R,, as in (b).] 2. Let x = f ( u , v ) , v = g(u, v ) be given as in Theorem I1 and let C,, enclose the origin 0. Show that the degree S can be evaluated by the formula

+

+

where x, y , d x , d y are expressed in terms of u , v:

(Hint: The line integral measures the change in 8 = arctan y / x , as shown in Section 5.6.) 3. Apply the formula of Problem 2 to evaluate the degree for the following mappings of the circle u2 + v 2 = 1 into the circle x2 + y2 = 1 : a ) x = - 3u , + 4v

Y = ~4u- - 3v

b) x = u2 - v 2 , y = 2uv c) X = u3 - u v 2 , y = 3u2v - v3

4

337

Chapter 5 Vector Integral Calculus

4. (Polar coordinates) Prove the validity of the transformation formula

+

$3

where R is the circular region x 2 y2 5 1 and x = r cos8, y = r sine. [Hint: Consider first the semicircular region R 1: x 2 y2 5 1, y 2 0, and the corresponding rectangle: 0 5 0 5 75, 0 5 r 5 1, in the r e plane. Show that the conditions of Theorem I1 are met, with S = 1. Note that the correspondence between the boundary of the rectangle and that of the semicircle is not one-to-one. Obtain a similar result for the semicircular region R2 : x 2 + y2 5 1 , y ( 0 and add the results for R 1 and R2.] 5. (The solid angle) Let S be a plane surface, oriented in accordance with a unit normal n. The solid angle C2 of S with respect to a point 0 not in S is defined as

+

R ( 0 , S) = k area of projection of S on SI,

i

I

3

+

where S1 is the sphere of radius 1 about 0 and the or - sign is chosen according to whether n points away from or toward the side of S on which 0 lies. This is suggested in Fig. 5.37. a) Show that if 0 lies in the plane of S but not in S, then Q ( 0 , S) = 0. flu b) Show that if S is a complete (that is, infinite) plane, then R ( 0 , S) = f2n. C) For a general oriented surface S the surface can be thought of as made up of small elements, each of which is approximately planar and has a normal n. Justify the following definition of element of solid angle for such a surface element:

r.n r3

dS2 = -do, where r is the vector from 0 to the element.

Figure 5.37

Solid angle.

Z

338

Advanced Calculus, Fifth Edition

d) On the basis of the formula of (c), one obtains as solid angle for a general oriented surface S the integral

Show that for surfaces in parametric form, if 0 is the origin,

Ix

z l

y

E 2 %

nco.s)=JJ

g 2

R,,,,

1

d u dv.

az (x2 + Y 2 + z 2 ) ~

This formula permits one to define a solid angle for complicated surfaces that interact , themselves. e) Show that if the normal of S1 is the outer one, then R ( 0 , S)'= 4n. f ) Show that if S forms the boundary of a bounded, closed, simply connected region R, then a(0, S), f4n,when 0 is inside S and R ( 0 , S) = 0 when 0 is outside S. g) If S is a fixed circular disk and 0 is variable, show that -2n 5 R ( 0 , S) 5 2n and that Q ( 0 , S) jumps by 437 as 0 crosses S. 6. (Degree of mapping of one sur$ace into another) Let S,,, and S,,, be surfaces forming the boundaries of regions R,,, and R,, respectively: it is assumed that R,,, and R,,, are bounded and closed and that R,,, is simply connected. Let S,, and Sxyzbe oriented by the outer normal. Let s, t be parameters for S,,,: u = U(S, t),

w = W(S,t),

u = U(S,t),

(a)

the normal having the direction of (u,i

+ usj + wsk) x (u,i 4 utj + w,k).

Let '

x = x(u, v, w),

z = z(u, v, w )

y = y(u, v, w ) ,

(b)

be functions defined and having continuous derivatives in a domain containing S,,,, and let these equations define a mapping of S,,, into S,,,. The degree S of this mapping is defined as 1/4n times the solid angle Q ( 0 , S) of the image S of S,,, with respect to a point 0 interior to S,,,. If 0 is the origin, the degree is hence given by the integral (Kronecker integral) X

Y

s 4=n ~ fJ $ R,,

?$

Z

t

2 3

1

(x2

+ y2 + Z2)t

ds d t ,

where x , y, 2 are expressed in terms of s, t by (a) and (b). It can be shown that 8 , as thus defined, is independent of the choice of the interior point 0 , that 6 is a positive or negative integer or zero, and that 6 does measure the effective number of times that S,,, is covered. Let S,,, be the sphere u = sins cos r , v = sins sin t, w = cos s, 0 5 s _( n,0 5 t 5 237. Let S,,, be the sphere x 2 + y2 + z2 = 1. Evaluate the degree for the following mappings of S,,, into Sxyz: a ) x = v , y=-w, z = u b) x = u 2 - v 2 , y =2uv, z = w J =

Chapter 5 Vector Integral Calculus

*5.15

PHYSICALAPPLICATIONS

2t

The following is a brief discussion of some of the important applications of the divergence and curl and of line and surface integrals.

a) Dynamics. If F is a force field, then, as shown previously, the work done by F on an arbitrary path C is W=

I

FTds.

(5.110)

C

In general, this will be dependent on the path. If, however, F is the gradient of a scalar, the work done will be independent of the path. If this holds, the scalar will be denoted by -U, so that

F = -grad U ,

U = U(x, y, z),

(5.111) and F is said to be derived from the potential U; U is also termed the potential energy of the force field. U is determined uniquely up to an additive constant: U =-

/

(x..v.z)

FT ds

+ const

(5.112)

(xl*yt.zl)

as in Section 5.6. In practice the constant is often chosen so that U approaches 0 as x 2 y2 z2 becomes infinite. The work done on a particle moving from A to B is then expressed in terms of U as follows:

+ +

W=

lB

f i d s = U(A) - ( I ( B ) ;

(5.113)

4 that is, the work done equals the loss in potential energy. The theorem work done equals gain in kinetic energy, proved in Section 5.4, holds for a general force field. Combining the two results, one concludes:

gain in kinetic energy = loss in potential energy (gain in kinetic energy)

+ (gain in potential energy) = 0,

''

since the gain in potential energy equals the negative of the loss. If one now defines the total energy of the particle to be E, where

E = (kinetic energy)

vL + (potential energy) = m+ U, 2

(5.114)

then one concludes that for an arbitrary motion of the particle under the given force, that is, the total energy is conserved. This is the law of conservation of energy for a particle. This was established here under the assumption that F was a gradient vector; it can be shown that such a conservation law can hold only

339

340

Advanced Calculus, Fifth Edition when F is a gradient vector. For this reason, force fields that are gradients are termed conservative force fields. b) Fluid dynamics. If u is the velocity vector and p is the density for a fluid motion, then, as in Section 5.11, the equation of continuity

a~ + div (pu) = 0 -

(5.116)

at

*e t?\ folds. This can also be written, by virtue of an identity of Section 3.4, as follows:

The first two terms are the Stokes total derivative of p:

and describe the rate of change of p as one stays with a particular particle of fluid in the motion. For an incompressible fluid, D p l Dt = 0, so that the equation of continuity becomes div u = 0,

(5.117)

and u is solenoidal. Another interpretation of div u and a new proof of the continuity equation are given in the problems that follow this section. The integral u~ d s about a closed curve C has already been introduced as the circulation of the velocity field. If this is zero for every closed path C , then by Theorem I11 of Section 5.13,

1,

r

.J.

curl u = 0.

(5.118)

If (5.1 18) holds, the flow is called irrotational. This implies, by Theorem I11 of Section 5.13, that the circulation is zero on every closed path, provided that attention is restricted to a simply connected domain D. If this last assumption is made, then u = grad 4 for some scalar 4, termed the velocity potential. If the flow is both irrotational and incompressible, then 4 must satisfy the equation div grad 4 = 0; that is,

r

and 4 is harmonic in D. c) Electromagnetism. An electromagnetic field is described, in accordance with Maxwell's theory, by two vector fields E and H, where E is the electric force and H is the magnetic field strength. Both E and H in general vary with time t .

Chapter 5 Vector Integral Calculus In the absence of conductors, E and H satisfy the Marwell equations:

a) div E = 47rp,

b) div H = 0,

(5.120) 1aH 1 aE d) curl H = -c ar ' c at' where p is the charge density and c is a universal constant. We now apply (a), . . . , (d) of (5.120) and our rules of vector calculus to show that E and H are expressiblein terms of suitablepotentials, which in turn satisfy certain partial differential equations. We assume that all derivatives appearing here are continuous, so that the order of differentiation can be changed and that the discussion is confined to a simply connected domain to which Theorem IV of Section 5.13 applies, so that each irrotational field is a gradient and each solenoidal field is a curl. From (b) of (5.120), we can choose a vector field A (the vectorpotential) such that

c) curlE =

H = curl A.

(5.121)

Here both H and A depend on x , y , z and t , as do all functions appearing in this analysis. As in Problem 5 following Section 5.13, A is not unique; to any choice of A one can add grad 4, where 4 is an arbitrary function of x , y , z and t . From (c) and (5.121) we deduce that .[SE 1 .? (5.122) and hence that Q (the scalar potential) can be chosen so that

If we take the divergence of both sides and use (a), we obtain

From (d) and (5.121) we obtain: 1 aE curl curl A - - - = 0. c at From (5.123), by differentiation,

T

*

a

(5.125)

Addition of the last two equations and application of the identity (3.40) of Section 3.6 for curl curl A gives

We now use the freedom in the choice of A to achieve the Lorentz condition:

341

I

342

Advanced Calculus, Fifth Edition

3

We return to this question below. Now from (5.124), (5.127) and (5.128) we obtain the desired partial differential equations for the scalar and vector potentials: 020-

1 a2@ = : --4

c2 a t 2 V ~ A --1 a 2 =~0 .

XP

(5.129)

7

~2

4

1

Ji

.3P

33?' r

d

(5.130)

at*

In each region free of charge ( p = 0), u = @ ( x , y , z , t ) satisfies the wave equation

Each component of A also satisfies the wave equation and hence from (5.123) so does each component of the electric force field in each region free of charge. We now consider the Lorenz condition (5.128). We can write A = A. grad Q,, where Q, is at our disposal. From A. and (5.122) we now can choose a scalar potential Q,o such that

+

From (5.123) and (5.132), 1 a ~ o @ 1- a A 1 a0 VQ,- VQo= -- -- - -- grad-. c at

at

C

at

Hence we satisfy (5.123) by taking

Substitution of this expression and that for A in (5.128) and rearrangement gives

Thus to achieve (5.128), we must choose Q, to satisfy this "non-homogeneous wave equation." This equation is studied in Chapter 10, as an equation for forced vibrations of a continuous medium; the general theory shows that solutions exist. Accordingly the Lorentz condition (5.128) can be satisfied. In the electrostatic case, H is taken to be 0 and E and @ are independent of t . Eqs. (5.123) and (5.124) give E = -V@, where Q, satisfies the Poisson equation,

v20= -47rp.

(5.134) 1

(5.135)

@ is termed the electrostatic potential. In any domain free of charge, @ is therefore a harmonic function.

4

343

Chapter 5 Vector Integral Calculus The function @ can be computed by Coulomb's law for given charge distributions. Thus for a point charge e at the origin, e =-

(5.136) + const, r = d x 2 + y2 z2. \ '! r and @ is obtained by simple addition for a sum of point charges. If the charge is distributed along a wire C and p, is the density (charge per unit length), .? then ,

+

@(XI, Y I ,

Z I )=

j 5+

I

conrt.

C

+

+

where rl = J(x - x ~ ) (y~- y1)2 (Z - z ~ )If~the . charge is spread out over a surface S, then @ is given by the surface integral

where p, is the charge density (charge per unit area) (see Section 5.18). ts;& d) Heat conduction. Let T(x, y , I , t) be the temperature at the point (x, y, z) of a body at time t. If heat is being conducted in the body, the flow of heat can be represented by a vector u such that the flux integral

.

S

for each oriented surface S represents the number of calories crossing S in the direction of the given normal per unit of time. The simplest law of thermal conduction postulates that

u = -k grad T,

(5.139)

with k > 0; k is usually treated as a constant. Equation (5.139) implies that heat flows in the direction of decreasing temperature and the rate of flow is proportional to the temperature gradient: /grad TI. I If S is a closed surface, forming the boundary of a region R in the body, : then . 4

by the Divergence theorem. Hence the total amount of heat entering R is

-

11

u,, do =

/fl

kdiv grad T d x d y d z .

(5.141)

R

S

On the other hand the rate at which heat is being absorbed per unit mass can also be measured by c 5 , where c is the specific heat; the rate at which R is receiving heat is then

10

c p z dx dy dz.

R

.

>

i.

344

Advanced~Calculus,Fifth Edition where p is the density. Equating the two expressions, one finds

10

- k div grad T

(cp

(5.143)

R

Since this must hold for an arbitrary solid region R, the function integrated (if continuous) must be zero everywhere. Hence aT - k div grad T = 0. at

cp-

(5.144)

This is the fundamental equation for heat conduction. If the body is in temperature equilibrium, aT/at = 0, and one concludes that div grad T = 0;

(5.145)

that is, T is hamtonic.

e) Thermodynamics. Let a certain volume V of a gas be given, enclosed in a container, subject to a pressure p. It is known from experiment that for each kind of gas there is an "equation of state"

connecting pressure, volume, and temperature T. For an "ideal gas" (low density and high temperature), Eq. (5.146) takes the special form: pV = R T ,

(5.147) .*

where R is constant (the same for all gases, if one mole of gas is used). With each gas is also associated a scalar U ,the total internal energy; this is analogous to the kinetic energy plus potential energy considered above. For each gas U is given as a definite function of the "state," hence of p and v: U = U(p, V).

(5.148)

The particular equation (5.148) depends on the gas considered. For an ideal gas, (5.148) takes the form:

where cv is constant, the specific heat at constant volume. A particularprocess gone through by a gas is a succession of changes in the state, so that p and V, and hence T and U ,become functions of time t. The state at time t can be represented by a point (p, V) on a pV diagram (Fig. 5.38) and the process by a curve C, with t as parameter. During such a process it is possible to measure the amount of heat, Q, received by the gas.

345

Chapter 5 Vector Integral Calculus

Figure 5.38

Thermodynamic processes.

The first law of thermodynamics is equivalent to the statement that

where Q(t) is the amount of heat received up to time t. Hence the amount of heat introduced in a particular process is given by an integral:

Now by (5.148), dU is expressible in terms of d p and dV:

Hence (5.151) can be written as a line integral:

or, with (5.152) understood, simply thus:

.

e

.

For (5.153) or (5.154) to be independent of the path C, one must have

"("1 av

ap

a au

-(-

=a p a v

+PI;

that is,

a2u --

avap

a2u

--

apav

+ 1.

"

'

;i

346

Advanced Calculus, Fifth Edition Since this is impossible (when U has continuous second derivatives), the heat introduced is dependent on the path. For a simple closed path C , , as in Fig. 5.38, the heat introduced is

*

~ 0; thus the heat introduced Since U is a given function of p and V, $ d = reduces to

J pdv. This integral is precisely the area integral l y dx considered in Section 5.5, with p replacing y and V replacing x . Hence for such a counterclockwise cycle the heat introduced is negative; there is a heat loss, equal to the area enclosed (a unit of area corresponding to a unit of energy). The integral l p d V can also be interpreted as the mechanical work done by the gas on the surrounding medium or as the negative of the work done on the gas by the surrounding medium. For the process of the curve C1 considered above, the heat loss equals the work done on the gas; the total energy remains unchanged, in agreement with the law of conservation of energy expressed by the first thermodynamic law (5.150). While the integral SdU p d V is not independent of path, it is an experimental law that the integral

.:

+

is independent of path. One can accordingly introduce a scalar S whose differential is the expression being integrated:

S is termed the entropy. In the first equation here, one can consider U and V as independent variables; in the second, p and V can be considered independent. Thus the first equation gives

and hence

a2s --

a2s

4 41 -2

3

Accordingly, one finds ap T-

2d a

ar

au - p -au + av = 0 aT

( U , V indep.).

Chapter 5 Vector Integral Calculus

347

A similar relation is obtainable from the second equation (5.155), and others are obtainable by varying the choice of independent variables. All these equations are simply different forms of the condition a P / a y = a Q / a x for independence of path (see Problem 8 following Section 2.1 1). The second law of thermodynamics states first the existence of the entropy S and second the fact that for any closed system,

that is, the entropy can never decrease.

PROBLEMS 1. a) A particle of mass m moves on a straight line, the x-axis, subject to a force -k2x. Find the potential energy and determine the law of conservation of energy for this motion. Does the law hold if a resistance -c% is added? b) A particle of mass rn moves in the xy-plane subject to a force F = -a2xi - b2yj.Find thc potential energy and determine the law of conservation of energy for this motion. 2. Let D bc a simply connected domain in the xy-plane and let w = ui - vj be the velocity vector of an irrotational incompressible flow in D. (This is the same as an irrotational incompressible flow in a 3-dimensional domain whose projection is D and for which the I-component of velocity is 0 whereas the x- and y-components of velocity are independent of 2.) Show that the following properties hold: a ) u and LI satisfy the Cauchy-Riemann equations: au- -av -

ax

b)

u and

C) j ' l r

ay'

a~ ay

-

av in D; ax

- - - A

.\.

.

,.

v are harmonic in D;

dx - v dy and I v dx

+ u dy are independent of the path in D;

d) there is a vector F = @i - $ j in D such that

e) div F = 0 and curl F = 0 in D;

f ) @ and $ are harmonic in D; g) grad @ = w, is constant on each stream line. The function @ is the velocity potential; $ is the stream function. 3. Let a wire occupying the line segment from (0, -c) to ( 0 ,c ) in thexy-plane have a constant charge density equal to p. Show that the electrostatic potential due to this wire at a point ( x i , yl) of the xy-plane is given by

+

where k is an arbitrary constant. Show that if k is chosen so that 0(1,0 ) = 0, then, as c becomes infinite, Q approaches the limiting value -2p log 1x1 I . This is the potential of an infinite wire with uniform charge.

I

I

I

348

Advanced Calculus, Fifth Edition

4. a) Write out in detail the steps leading from the Maxwell equations (5.120) to the partial differential equations (5.129), (5.130). b) Show from Eqs. (5.120), . . . , (5.130) that each component of E satisfies the wave equation in each domain free of charge. 5. Find the temperature distribution in a solid whose boundaries are two parallel planes, d units apart, kept at temperatures T I ,T2,respectively. (Hint: Take the boundaries to be the planes x = 0 , x = d and note that, by symmetry, T must be independent of y and z.) 6. Show that on the basis of the laws of thermodynamics, the line integral

1

is independent of the path in the T V plane. The integrand is minus the differential of the free energy F.

7. Consider a fluid motion in space. A particle occupying position (xo,yo. zo) at time t = 0 occupies position ( x , y , z ) at time t . Thus x , y , z become functions of xo, yo, 20, t : x = ~ ( x oYO, , 20, t ) , y = @(xo,yo* ZO. t ) , z = x(x0, yo, zo, t ) .

2

(*)

Let the V symbol be used as follows:

and let J denote the Jacobian

41

a ( x , Y ,Z ) a(xo7yo, Z O ) . Let v denote the velocity vector: ax ay az a4 v=-i+-j+-k=-i+-j+-k. at at at at a) Show that J = V x . V y x V z . b) Show that axo = i . -

vyx vz

ay0 = j . v y -

ax J ' ax and obtain similar expressions for

C)

a@

ax

at

at

x

vz

J

'

az0 = k . V y x ax

J

vz '

(Hint: See Problem 5 following Section 2.12). Show that

aJ --=vv,.vy at d) Show that

xvz+vv,.vzxVx+vv,.vxxvy.

1 aJ divv = --. J at

4

Chapter 5 Vector Integral Calculus

Figure 5.39

349

I

Problem 7.

[Hint: By the chain rule,

I

Use the result of (b) to show that

and obtain similar expressions for av,/ay, av,/az. Add the results and use the result of (~1.1 Remark The Jacobian J can be interpreted as the ratio of the volume occupied by a small piece of the fluid at time t to the volume occupied by this piece when t = 0,as in Fig. 5.39. Hence by (d) the divergence of the velocity vector can be interpreted as measuring the percentage of change in this ratio per unit time or simply as the rate of change of volume per unit volume of the moving piece of fluid.

8. Consider a piece of the fluid of Problem 7 (not necessarily a "small" piece) occupying a region R = R(t) at time t and a region Ro = R(0) when t = 0. Let F(x, y , z, t) be a function differentiable throughout the part of space concerned. a) Show that FW(XO,YO,ZO. t ) . . . .I J

JJJ~(x,y.Z,t)dxd~d~= R(0

I YO ~ Z O .

~ X O

Ro

[Hint: Use Eq. (5.109), noting that the degree 6 must be 1 here.] b) Show that

d dt

/// R(0

F(x, y, z, t ) d x dy dz =

/I[$ /+ R(t)

I

div(Fv) d x dy dz.

[Hint: Use (a) and apply Leibnitz's Rule of Section 4.9 to differentiate the right-hand side. Use the result of part (d) of Problem 7 to simplify the result. Then return to the original variables by (a) again.] 9. Let p = p(x, y, z , t) be the density of the fluid motion of Problems 7 and 8. The integral

4 350

I

1

Advanced Calculus, Fifth Edition represents the mass of the fluid filling R ( t ) . The conservation of mass implies that this integral is constant:

Use this result and that of Problem 8(b) to establish the continuify equation

[Hint: Cf. the derivation of the heat equation (5.144).]

"5.16 POTENTIAL THEORY IN THE PLANE In this section and the following section we introduce potential theory in space of two or three dimensions. The 3-dimensional theory reduces to the 2-dimensional one if one considers mass or charge distributions in xyz-space that do not depend on z and modifies the potential. The process is illustrated in Problem 3 following Section 5.15. The Newtonian potential m l r in space turns into a logarithmic potential m log l l r in the plane. We consider a force field F in the xy-plane given by the equation

4

+

where r = P Q , r = Irl, Q is fixed at ({, r ] ) , P is (x, y ) , a variable point, k and m are constants, k > 0. Thus F is directed from P to Q and hence is an attractive force, if m > 0. Its magnitude is

Thus the force varies as the inversejrst power of the distance. The force F is the gradient of a scalar U = U (x, y):

For r = [(x - t)'

+ (y - r ] ) 2 ] 1 / 2 and hence aU - k m a r km = -(6

ax

so that, as asserted, grad U =

r ax

, [(t km r

- x)i

r2

-XI,

+ (v - y)j] = -km ~ =rF. r

One calls U = I J ( . t . , v ) the logarithmic potential due to the point mass m at Q and unit mass at P.

2 '

" U

..

Chapter 5 Vector Integral Calculus

351

By proper choice of units, one can assume that k = 1, and we shall do so. Thus

We could add an arbitrary constant to U and still have grad U = F, but we shall not do so. For a finite set of particles of masses mi at Q i(i = 1, . . . , n), one has an associated force field

and as previously, F = grad U,

here U is the logarithmic potential due to the finite set of masses. By passage to the limit, one obtains logarithmic potentials of other mass distributions. For mass spread over a region E with density ~ ( 6q), the potential is U(x,Y ) =

I/

p log

1

;d5 d v ,

E

+

r = J(6 - x ) ~ (v - Y ) ~ .

(5.163)

The corresponding force is

+

Here V,, = ( a / a x ) i ( a / a y ) j .The vector integral is interpreted, as in Section 4.5, as

In (5.163) we assume that the density p ( 6 , r ] ) is continuous in E and E is a bounded closed region, as is usual for double integrals. If P: ( x , y) is outside E, then l l r is continuous in (6, q ) on E, so that U ( x , y) is well defined. If P is inside E or on the boundary of E, then the integral in (5.163)is improper. However, as in Section 4.8, the integral does exist. For example, if p ( 6 , q ) does not change sign, then as in Section 4.8, one is led to consider a limit process at (x, y). We use polar coordinates R, B with origin at ( x , y ) , so that R = r . In these coordinates, one must consider lim

h+O+

/2R

0

1 R

L o p ( 6 , )1)1og-RdRdB,

,

352

Advanced Calculus, Fifth Edition

Figure 5.'d7Formation of double layer.

and, since R log 1/ R + 0 as R + 0, the limit does exist. The conclusion also holds for p of variable sign (as occurs in the electrostatic case) (see Section 6.26). Obtaining F from U requires differentiating under the integral sign. For P outside E, Eq. (5.164) follows from Leibnitz's Rule of Section 4.9. For P in E the result is correct, but a more careful analysis is needed. One can in fact show that grad U is given by (5.164) for all ( x , y) and both U and grad U are continuous everywhere. One can also, by a passage to the limit from (5.162), obtain the logarithmic potential of a distribution of mass on a curve C . One is led to a line integral

4

Here we find that if C is piecewise smooth and p is continuous, then U is continuous everywhere. However, the corresponding force field grad U is defined only for P off of C. Another limit process starts with two "parallel" curves C,, C2, close together with mass (or charge) densities equal but opposite in sign at adjacent points, as suggested in Fig. 5.40. If the two curves are brought to coincidence while the densities become infinite in the proper manner, one obtains as limit a new potential U ( X ,Y)=

1

a 1 Y(s)-an log r ds,

(5.167)

C

called the logarithmic potential of a double layer on C. In electrostatics this corresponds to a dipole layer on C . The curve C is assumed to have a continuously varying unit normal vector n, with respect to which the directional derivative alan is formed; y(s) is assumed to be continuous on C. By contrast to (5.167) the potential (5.166) is called the potential of a single layer on C. For (5.167), one finds that U itself is discontinuous on C. However, for P off of C, U is well defined and continuous, and Leibnitz's Rule can be applied to form the force field F, which is also continuous off of C. EXAMPLE The logarithmic potential of a mass distribution of constant density b over a disk of radius a with center at (0,O) is given by 'I= b

lalLl

log

1

R2

+ p2 - 2Rp C O S -( ~

p dB dp.

(5.168)

ff)

Here x = R cos a,y = R sin a , and we used the law of cosines to express r in terms of R, p , 8 , a , as in Fig. 5.41. With the help of advanced integration formulas,

Chapter 5 Vector Integral Calculus

. -

1

Figure 5.41

Distances and angles for Eq. (5.168).

Figure 5.42

Potential and force magnitude for Eq. (5.168).

one finds: %Ha2 - R2 - 2a2 logal, R 5 a IT ba2 log

R>a

R,

(see Problem 1 following Section 5.18). At R = a the two expressions give U = nba2log 1/ a so that U is continuous for all ( x , y). Also

=

(

-nb(xi nbu2

--jir(xi

+ yj),

+ yj),

R 5a R > a,

(5.170)

from which we see that F is continuous everywhere. The force is radial everywhere, directed toward the origin. The force magnitude depends only on R ; this magnitude and U are graphed as functions of R in Fig. 5.42.

U as a harmonic function. (x, y) for ( x , y) not at

For fixed

(t, rl). For

(6, q) the function log l / r is harmonic in

and the sum is 0. Leibnitz's Rule permits a similar differentiation under the integral sign for the potentials (5.163), (5.166), and (5.167). Hence we conclude that away from the masses U ( x , y) is harmonic. Equivalently,

div F = div grad U = 0.

353

I

4354

Advanced Calculus, Fifth Edition This is illustrated by the example of Eq. (5.168), for which U is a constant times !? log 1/R (R2 = x2 y2),away from the mass. B

+

R

Behavior of U and Ffor large R. The results to be presented hold for all types of mass distributions, but we formulate them only for the case of a continuous mass distribution on E, as in (5.163). We ask how U and F behave when P: ( x , y) recedes to infinite distance, that is, when R + oo,R = (x2 y2)1/2.We let

+

E

we interpret M as the total mass. For U itself we have the following assertion: For large R , R 2 Ro,

where p(x, y) is bounded, Ip(x, y)l 5 const. Thus for large R the potential behaves as if the mass were concentrated at the origin, with a small error, which approaches O a s R - + oo. To prove this, we led d be so large that E is included in the circular region x 2 y 2 j d2. Then by the triangle inequality (a b 2 c for a triangle of sides a , b, c),

+

+

R ~ r + d ,r ~ R + d (see Fig. 5.43). Therefore for R > d,

,. so that

Figure 5.43

Derivation of Eq. (5.172).

Chapter 5 Vector Integral Calculus

In particular, R / r 2 1/2 for R 2 Ro = 3d: Next by (5.171),

'

Now for x > 0.5, I log x 1 i 12(x - 1)1, as one sees from a graph of y = log x and y = 2(x - 1) (Fig. 5.43). Therefore by (5.173), for R 2 Ro,

The last expression is continuous for R 1 Ro = 3d and has limit const. d as R + oo. Therefore p(x, y ) is bounded, as was asserted. Next we turn to F = grad U . From the preceding result we expect that for large R, F is approximately equal to the gradient of a potential due to mass M at the origin; that is, for large R, 1 M F -- grad M log - = --(xi R R2

M + yj) = --G. R2

Thus F should be approximately purely radial for large R with magnitude M I R. Now we introduce the unit vectors

which we term the unit radial and tangential vectors, respectively (Fig. 5.43). As the figure shows, we have corresponding components of F: Frad=F.u,

Fta,=F.V.

We can now assert: For R 2 Ro,

where Iq(x, y)l 5 const and Is(x, y)l 5 const. The proofs are left as exercises (Problem 5 following Section 5.18). Mass relationship. We know that div F = 0 away from the masses. It follows from the Divergence theorem that

'

355

356

Advanced Calculus, Fifth Edition

I

for every simple closed path C in a simply connected domain containing no mass. Here n is the exterior normal with respect to the region enclosed by C. If C does enclose mass, then this line integral will generally not be 0.For example, for the case of one particle of mass m at the origin, m F = --R (R = xi yj),

+

R2

and hence by Eq. (5.39),

for every path C enclosing the origin (Example 2 in Section 5.6). As in Section 5.7 (see especially Problem 8 following that section), if C encloses a finite set of masses m 1, . . . , m k ,then

#

-

F - n d s = -2n(mI

+

.

.

a

+

mk).

-.

C

In general, we are led to assert: For each mass distribution the gradient F of the corresponding logarithmic potential has the property that

-$ F .

n ds = 2n x total mass enclosed by C.

(5.176)

C

This conclusion is valid, provided that C does not pass through a point at which a mass particle is located or which lies on a curve on which there is a single or double layer. For the case of a distribution over a region E we first take the case in which C completely encloses E. We then observe that the value of

c

is the same for all such C. For since the integral is 0 for all curves enclosing no mass, we are dealing with a line integral P dx Q dy for which aQ/ax = a P / a y wherever there is no mass. Thus as in Section 5.7, the integral has the same value for all closed paths C enclosing the same "holes." To evaluate the integral, we take C to be a circle x 2 y2 = a2,where a is large. Then along C, F . n is simply the radial component of F. By (5.174),

+

+

where Iq(x, y)l 5 K = const for a sufficiently large. Hence

4

Chapter 5 Vector Integral Calculus

Figure 5.44

Proof of mass relation.

.

The last term is in absolute value at most

,

-

Jr

and hence it can be made as small as desired by taking a sufficiently large. But the value of the integral (5.177) is independent of a . Thus the second term must be 0. This gives

I

F . n ds = -217 M = -2n (total mass).

C

If C encloses only some of the mass, as in Fig. 5.44, then we can write F =

F,,,+F,,,, where F,,, is the force field arising from the mass inside C and F,,, is that arising from the mass outside C. For F,,, there is now no mass inside C , so that

f

F,,, . n d s = 0 .

C

For F,,, we conclude as previously that

F,,, . n ds = -2n(mass inside C). Hence for F = Fin,+ FeXt,

6F .

n ds = -2n(mass inside C).

Poisson equation. For the mass distribution just considered, let us take a point (xo, yo) interior to E and take C to be a circle Cg of radius 6 about this point, bounding a circular region Eg . Then we can write

For the integral on the right-hand side gives the total mass inside C g .We divide both sides by ns2, the area of Eb:

if F . n d s = -2n ns2 C

17s2 Es

357

358

Advanced Calculus, Fifth Edition We now let 6 + 0. As in Section 5.11, the left-hand side approaches div F, evaluated at (XO,yo); the right-hand side approaches -2np(xo, yo), as in Section 4.3 [Eq. (4.49)]. Therefore for every point (x, y) interior to E, div F = - 2 n j ~ ( x ,y).

(5.178)

Since F = grad U ,we have also

j

div grad U = V ~ = U -2np(x, y)

(5.179)

at every such point. Equation (5.179) is called the Poisson equation. The replacement of div grad U by V ~ in U (5.179) assumes that the potential U has continuous second partial derivatives at (x, y). This can be shown to be valid if, for example, p(x, y) has continuous first derivatives. (See Chapter 5 of the book by Sternberg and Smith listed at the end of the chapter.)

Remark.

The first two identities of Green are given in Problems 11 and 12 following Section 5.7. The third identity is the equation

4

Here u is assumed to have continuous first partial derivatives in a domain containing the closed region E bounded by the curve C and to be harmonic inside C. The point (x, y) is interior to C, n is the exterior normal on C , and r is as above the distance from a general point r ] ) on C to (x, y). We remark that the term

(c,

is a logarithmic potential of a single layer on C and the term

is the logarithmic potential of a double layer on C. Hence the identity asserts that a harmonic function can be represented as the sum of two such logarithmic potentials. To prove the identity, one starts with the second identity of Green:

(Problem 12(a) following Section 5.7). We apply this first to our given region E, taking v = log l / r , where r is as above the distance from r ] ) to (x, y), but take

(c,

I

359

Chapter 5 Vector Integral Calculus (x, J J ) exterior to E . Then v is harmonic as a function of (c, g) in E , so that (5.181) becomes

SS t

au

log -V'U dc dg =

E

- log C

an

1 -

r

a

- u - log an

(5.182)

If, in addition, u is harmonic in E, then (5.182) becomes

This is valid for (x, y) outside of E. Now for (x, y) interior to E we draw a small circle Cgof radius 6 , with center at (x, y), and let Eg be the region obtained from E by deleting the interior of this circle from E (Fig. 5.45). We again take v = log 1/ r and can apply Eq. (5.181) to the region E g ,since (x, y) is exterior to this region. We obtain

On Csthe normal is exterior to E g ,SO that a / a n = -a/&. Thus in polar coordinates r, 8, with center at (x, y), so that ds = -6 dB on Cs,traced clockwise,

Figure 5.45

Proof of third Green identity.

r

!

360

Advanced Calculus, Fifth Edition We assume that u has continuous partial derivatives in E, from which it follows that, as S + 0 , the last integral has a finite limit. But S log 116 + 0 as S + 0 . Thus

Next

fu

i log d s = -

c6

6

:Ir=8

u(6,O)- log :r

( - 8 ) dB

*

As S + 0, this has limit

u(0,O)dO = 2 7 5 ~ at ( x , y). Accordingly, as 6 + 0, the double integral over Eg has a limit, which is just the integral over E, as is usual for logarithmic potentials. Thus

au 1 a JJ log : ~ ' ud l dr] = f (log - - u- log an r an r E

C

'1

ds - 2nu(x, y ) .

(5.184)

If, finally, u is harmonic in the interior of E, then the left-hand side is 0 , and the identity (5.180) is proved. Remark. The proof applies to a general region E bounded by several closed curves C, , . . . , Ck,SO that as for the first and second identities, we can replace the integral over C in (5.180)by one over BE, properly directed.

Green'sfunction. The identity (5.180) gives the harmonic function u in the interior of E in terms of the values of u and aulan on the boundary. One can show that the interior values of u can be found from those of u alone on the boundary. There is a corresponding formula

using a Green's function G , a function of x , y , 6 , and r] having special properties. For the case of a circle of radius a , Eq. (5.185)can be written as follows:

Here R = (x2

+ y 2 ) ' / 2 .This is the Poisson integral formula for the circle.

d 7

4

:

Chapter 5 Vector Integral Calculus For further details on logarithmic potentials, see the books by Kellogg and Sternberg and Smith listed at the end of the chapter.

There is a potential theory for 3-dimensional space analogous to the theory of Sections 5.16 and 5.17 for the plane. In space the force law is that of gravitation. Hence for a single particle of mass rn at Q, for appropriate choice of units,

+

where as before r = PQ, Q is fixed at (t, q, 0, P is at (x, y, z), r = Irl. The force magnitude is, for rn > 0,

We verify that

We call U(x, y, z) the Newtonian potential due to the single mass rn at Q. We generalize as in Eqs. (5.161) and (5.162) to several particles and by passage to the limit to the distribution of mass over a region with density p(6, q , 5 ) :

One verifies as in Section 5.16 that the improper integrals exist and that U and F are continuous in all of 3-dimensional space (see Section 6.27). One has also single- and double-layer potentials:

u=

11)(;.

do.

S

In (5.192), S must have a continuous unit normal vector field n. Again U is found to be harmonic away from the mass: div F = div grad U = V ~ =U0.

(5.193)

361

362

Advanced Calculus, Fifth Edition 1

+ +

For large R = ( x 2 y2 z2)'I2,again U and F behave as if all the mass M were concentrated at the origin ( 0 , 0 , O ) : (For 5.190),

for R sufficiently large, where Ip(x, y , z)l and Iq(x, y , z)l are bounded. Thus the force field is mainly radial for large R , and one finds as in the plane that -

//

F . n do = 4n . mass enclosed by S.

(5.196)

S

Here S is the boundary of a bounded closed region E, and n is the unit exterior normal on S. From (5.196),one concludes as in the plane that for a distribution over a solid region E with continuous density p , one has = -4nl(t, div F = V ~ U

r ] , {)

(5.197)

at each interior point of E. Green's third identity generalizes to space in the form

where S bounds E, u has continuous first partial derivatives in a domain containing E, ( x , y , z ) is interior to E, and u ( x , y , z ) is harmonic on the interior. There is also a formula analogous to (5.185) for space, using a Green's function. For the sphere S: x 2 y2 z2 = a 2 this becomes a Poisson integral formula:

+ +

+ + + +

On the right, R' = x 2 y2 z 2 , u is evaluated at points (6, 17, {) on S , a is the angle between the vectors xi y j z k , and t i q j {k, ( x , y , z) is a point interior to S . For details, one is referred to the books of Kellogg and Sternberg and Smith listed at the end of the chapter.

+ +

Helmholtz's theorem. Let G be a vector field whose components G,, G , , G, are continuous over the bounded closed region E. Then G=VxH+V@, for appropriate H and 4, on the interior of E. To prove this, we let

(5.200)

1

Chapter 5 Vector Integral Calculus

363

so that the component of U are Newtonian potentials. By applying (5.197) to each component of U we conclude that

V'U = -4nG

(5.202)

on the interior of E (see Eq. (3.39) in Section 3.6). By Eq. (3.38) in Section 3.6,

V ~ U = -curl curl U + grad div U, so that by (5.202),

1 G = -curl 4n

curl U

1 4rr

- -grad

div U,

and this gives (5.200), with H = (1/4n) curl U, q5 = -(1/4n) div U. Thus the proof is complete. One can extend the result to unbounded regions, provided that the corresponding improper integrals converge.

PROBLEMS 1. Use the result of Problem 6 following Section 4.9:

I"

log(1 - c cos 0) d0 = 217 log

1 + m , Osctl, 2

to evaluate the logarithmic potential U of the example in Section 5.16. . 2. Another way to evaluate the potential U of the example in Section 5.16 is to use the following information: a) V'U = -2rrb for R < a , b) V ~ = U0 for R > a ,

%,.

k +9

C) U = M log for large R, as in Eq. (5.172), d ) U and grad U are continuous for all (x, y), e) U depends on R alone, by symmetry. Since V'U = a 2 ~ /R~ a + ( 1 / R )a U / a R in polar coordinates, for a function depending only on R (Section 2.17), (a) and (b) give differential equations for U . By (c) and (d) the arbitrary constants in the solution can be determined. Carry out the process suggested to find U . 3. Let U be the logarithmic potential of a distribution of mass with constant density b on the circle x2 y2 = a 2 . a) Find U by integrating, with the aid of the formula given in Problem 1. b) Find U as in Problem 2, omitting (a) and replacing (d) by the condition that U is ti . continuous for all (x, y) and (b) by v 2 u = 0 for R # a .

+

c) Find grad U and show that it is discontinuous for R = a . 4. ~ eUt be the logarithmic potential of a double layer of constant moment y on the circle x 2 y2 = a 2 ,with n the outer normal.

+

."

364

Advanced Calculus, Fifth Edition a) Find U by integration. [Hint: To find (a/an)log l / r , write r 2 = p2 + R~ 2 p R cos(6 - a ) , where p , 8 are the polar coordinates of (6, q ) and R , a are those of (x, y ) . Then a/an = slap, to be evaluated for p = a.] b) Find grad U and show that the normal derivatives of U have a discontinuity on the circle. 5. a) Prove (5.174) with the conditions stated. [Hint: Use (5.164) to obtain Frad= F . U. where u = cosei + sin 8 j = (xi + yj)/R. Show by (5.174) that q can be written as

Show by (5.173) that each of the terms inside the brackets is bounded by a constant for R > ro = 2d.l b) Prove (5.175) with the conditions stated. [Hint: Show by (5.175) that

and use (5.173) to show that the integrand is bounded for R 2 Ro = 2d.l 6. Find the Newtonian otential U of the solid sphere x 2 y2 z2 5 a 2 with density p = k J m p V e r i f y that V'U = - 4 n p inside the sphere. 7. Show that for a mass distribution of constant density on a spherical surface there is no net force on a particle inside the surface. 8. Justify Eq. (5.194) for the Newtonian potential (5.190), with continuous density p , where p ( x . y . Z ) is bounded for R large.

+ +

+

The expressions occumng in our integrals: f (x) d x , F ( x , y) d x dy , P d x Q dy , L dy dz M dz dx N d.r dy, . . . are termed differential forms. The mathematician lie Cartan (1869-195 1) developed procedures for working with such forms which yield our rules and many others. We here outline the principal ideas, giving references for a full treatment. For this discussion it is convenient to generally regard the integrals as line or surface integrals or their generalizations to higher dimensions. Thus

+

+

lb

f (x) d x and

11

~ ( xy), dx d y

S

are regarded, respectively, as a line integral of f on the path from a to b and as a surface integral of N over an oriented surface S in the xy-plane, as in Fig. 5.28 in Section 5.10. We even distinguish between d x dy and dy dx. The order dx dy is considered to be associated with the normal orientation of the xy-plane and, if S has that orientation, j i s d~x dy has the value j J s N d A as in Chapter 4. If S has the opposite orientation, this value is multiplied by - 1. In either case

Chapter 5 Vector Integral Calculus Similar conventions apply to triple integrals JIJ f (x, y , z)dx dy dz and to multiple integrals in general. In En the coordinates x ' , . . . , xn or the corresponding basis vectors el = ( 1 , 0 , . . . , O), . . . , e n = (0, . . . , 0 , 1) of V n (Section 1.14) in that order will be considered to provide the positive orientation; any interchange of two coordinates will reverse the orientation, as will any odd number of interchanges (but not an even number of interchanges). Each region R of integration is assumed oriented, either positively or negatively, and

where the integral on the right is the usual integral with respect to n-dimensional volume over the unoriented region R; the sign holds if i,, . . . , i,, is an even permutation of 1, . . . , n and R is positively oriented; replacing "even" by "odd or reversing the orientation of R multiply the value by - 1. (See Section 1.4 and Problem 16 (d) following Section 1.5.) From these conventions we are led to our first rule for differential forms:

+

fdxdy= -fdydx and in general

+

with or - accordingly as ( j l , . . . , j,) is an even or odd permutation of i l , . . . , I', . We clearly have no use for dx dx and we write fdxdx = O and, in general

f dx" . . . dxi0'= 0 if two of i l . . . . , im are equal.

(5.204)

Here 0 is a zero form, which can be identified with Odx' . . . dxm. We now define a dflerenrial form of degree m, or an rn-form, in n variableitto be a sum of terms of form c.

where A is a function of x ' , . . . , x n , 1 5 il 5 n, . . . , 1 ( i, ( n. We also allow m = 0 and consider as 0-form a function ~ ( x, .' . . , xn). All functions will be assumed to be differentiable as required in an appropriate open region. Two m-forms are considered to be equal if both can be reduced to the same form by applying the rules (5.203), (5.204), and combining terms of the same indices ; I , . . . , i,. Thus each m-form can be written as a sum of terms (5.205) with no two ~ndicesequal and no two terms having the same set of indices-for example, each sum can be written in exactly one way as a sum

We remark that as a special case, a form of degree n can be written in just one May as

365

366

Advanced Calculus, Fifth Edition

1 1

and hence thought of as B dV, where V is n-dimensional volume. Further, a form of degree m > n can be written uniquely as a zero form, as follows from (5.204), since for m > n, the il, . . . , i, cannot be all distinct. We can now perform the algebraic operations of addition and multiplication of m-forms, for given n , in the obvious way, using the rules (5.203) and (5.204) to simplify, combining terms where possible. For addition we combine only terms of the same degree m and obtain results of this degree. The multiplication, called exterior product, combines two forms of degrees m ~m2 , to obtain one of degree ml m 2 (reducing to a zero form if ml m2 > n). One verifies that these definitions respect the definition of equality of forms given above: that is, replacement of a, B by equal forms crl , @ I , replaces the sum or product y by an equal form yl (see Problem 2 following Section 5.22).

+

+

Remark. The term (5.205) is sometimes written as Ai,...i, dxil A dxiz. . . A dxitn and exterior products (also called wedge products) are also written with a A. EXAMPLE 1

With n = 5

One now verifies algebraic laws:

Here a, j l and y are forms of degrees ml , m2, m3 respectively and in (5.206) ml = m2 = m3, in (5.208) m2 = m3. EXAMPLE 2

We illustrate the first rule (5.207):.

(A dx' dx3dx5)(lIdx2dx4) = AB dxl dx3dx5dx2dx4,

(B dx2d x 4 ) (dx' ~ dx3dx5) = BA dx2dx4dxl dx3dx5. Here A B = BA (scalar multiplication) and >liJ

= -dx2 dx' dx3dx5dx4 =

dx2dxl dx3dx4dx5

367

Chapter 5 Vector Integral Calculus thus six interchanges were used to go from one order of the d x i to the other and mlmz = 3 x 2 = 6, so the rule is satisfied. For the general case, see Problem 3 following Section 5.22. Now we introduce a derivative, the exterior derivative. If w is an m-form (here usually m < n), the exterior derivative of w is an (m + 1)-form, denoted by dw. If w is a single term (5.205), then

For m = 0, w = A, a scalar function, and aA dw = -dx' ax1

aA + ...+ dxn axn

as usual. For o a sum of terms, dw is obtained by applying (5.209) to each term and adding. The results can be simplified by the rules given above. From the definition just given one deduces further rules. For an m 1-form a! and m2-form B, d ( a + B ) = d a +dB d(@) = (da)B

(here ml = m2),

(5.211)

+ (-l)m'a(dS),

(5.212)

d(da) = 0.

(5.213)

(See Problem 4 following Section 5.22.)

EXAMPLE 3

Let n = 3 and w = A , d x l

+ A2 d x 2 + A3 d x 3 .Then

+ A2j + A3k. Let n = 3 and w = A' d x 2d x 3 + A2 d x 3d x l + d x l d x 2 . hen

In familiar notation, this is just curl v .d a , where v = A l i

EXAMPLE 4

In familiar notation, this is div v d x dy dz.

~3

@

368

Advanced Calculus, Fifth Edition >

*5.20 CHANGE OF VARIABLES IN AN m-FORM AND GENERAL STOKES'S THEOREM If one is given equations describing a differentiable mapping from a region in EP into our given region of Section 5.19 in E n ,then each m-form in x ' , . . . , x n is automatically converted into an m-form in y ' , . . . , yP by expressing each scalar function in terms of the new variables and applying the rules in Section 5.19. EXAMPLE 1 The 2-form w of Example 4 of the preceding section becomes, after the substitution X

i

= xi(u, v),

a ( ~x ~~ ,)

i = l , 2 , 3,

(5.214)

+ A2(. . .) a(x3, + A3(. . a(u, V ) XI)

du dv.

a(u, v) a(u, V ) This is exactly the procedure used in evaluating the surface integral of our 2-form. Thus

as in Eq. (5.81) in Section 5.10. Here we are assuming that the uv-plane has its usual orientation, that R,, is positively oriented and that the orientation of S is determined by that of R,, by the mapping equations (5.214)-that is, that the sign holds in Eq. (5.81).

+

A similar discussion applies to other substitutions.

EXAMPLE 2

A line integral (integral of a I-form) Pdx J

+ Qdy

j

in which C is given by x = f ( t ) ,y = g(t), as in Section 5.2, becomes

as in Eqs. (5.4) and (5.5). Here we assume that C has the orientation (direction) given by increasing t.

Chapter 5 Vector Integral Calculus EXAMPLE 3

A double integral

can be considered as the integral of a 2-form F(x, y) dx dy over the positively oriented region R,,. The substitution x = x(u, v), y = y(u, v), as in Example 1, gives the integral

over the oriented region R,, in the uv-plane (in which du dv is associated with the positive orientation). The two double integrals are equal if the mapping is one-toone and preserves orientation-that is, if a(x , y)/a(u, v) > 0; they are also equal if a(x, y)/a(u, v) < 0, for then R,, is negatively oriented and there are two changes of sign. Equality also holds under more general conditions, as in Section 5.14. In general, one verifies in the same way that, under a differentiable mapping

with differentiable inverse, from one open region in En to another, each line, surface or hypersurface or multiple integral of an m-form over an appropriate curve, surface etc. is converted by the change of variables procedure into a new integral of an m-form and the value of the integral is unchanged. General Stokes's theorem. Green's theorem, the 2- and 3-dimensional Divergence theorems and Stokes's theorem can be considered as special cases of one theorem, ' called the general Stokes's theorem:

Here w is a general m-form in n-dimensional space, m < n, and S is a hypersurface of dimension m + l((m + 1)-dimensional manifold) with boundary B, an m-dimensional hypersurface; S and B are consistently oriented (as suggested by our rule for Stokes's theorem in Section 5.12). EXAMPLE 4 Let m = 1, n = 2, let B be the simple closed curve C bounding a ' positively oriented plane region R in the xy-plane, as in Section 5.5. Then (5.215) becomes

Since d ( P dx + Q dy) = (aQ/ax - aP/ay) dx dy as in Example 3 in Section 5.19, we recover Green's theorem and hence, as in Section 5.5, the Zdimensional Divergence theorem. •

369

370

IB

Advanced Calculus, Fifth Edition

1

Examples 3 and 4 of Section 5.19 show how Eq. (5.215) leads to the familiar Stokes's theorem and Gauss theorem. The proof of the general theorem is given in the books by Boothby (pages 25 1-258), Hodge (pages 75-78) and Flanders (pages 64-66), listed at the end of the chapter. These three books are valuable general references for differential forms.

Let an m-form (m 2 I ) be given in En.We can write it as

where we follow the summation convention of Section 3.9, summing over the repeated indices i l , i2, . . .., each going from 1 to n. Thus a general 2-form in E 2 is

By the rules of Section 5.19, the m-form can be written in many different ways. In particular, we can write it (in precisely one way) as in (5.216), with A,, ,,,, = 0 if

ip = iq for some p # q ,

(5.217)

AJI J , = fA'] 1," (5.218) if ( j l , . . . , j,) is a permutation of distinct integers i l , . . . , i,, with (-) if the permutation is even (odd). (See Problem 7 following Section 5.22.) For the rest of this section we shall assume that all our forms are written in this way. We further assume as before that the coefficients are differentiableas required in an open region D of E n . At a chosen point of D, for given numerical choices of d x l , . . , dxn,the form (5.216) has a numerical value. Hence, as a tensor, it should have order 0, or be an invariant. Thus, as in Section 3.9, to show that we have a tensor, we must show the values are unaffected by a change of coordinates

+

.

x ' = x l ( ~,..., l fn), i = 1 , ..., n. The d x i are the components of a contravariant vector since .

(5.219)

axi ax]

dx' = -d.fJ.

From our discussion of the Ail they appear to be the components of an alternating covariant tensor as in Section 3.11. To verify this, we make the substitution (5.219). As in Section 5.20 the given form becomes

Thus

Chapter 5 Vector Integral Calculus This shows that Ai, ,,,in, is indeed a covariant tensor of order m. As in Section 3.11, it is alternating in all allowed coordinate systems. [We note that the fact that the integral of an m-form does not change its value under a change of coordinates, as pointed out in Section 5.21, also implies that the m-form is an invariant.] Thus an m-form is an invariant obtained by contraction of a mixed tensor * $4

1

(setting jl = il, . . . , j,,, = i!, and summing as in Section 3.9). The result is an invariant, just as d s 2 = g,j dx' d x j is an invariant obtained by contraction. One verifies that the exterior product of alternating covariant tensors, as defined in Section 3.1 1 , is equivalent to the exterior product of differential forms. (See Problem 8 following Section 5.22). Finally formation of the exterior derivative is a valid tensor operation: that is, validity of the relation B = d a in one allowed coordinate system implies its validity in all allowed coordinate systems. (See Problem 10 following Section 5.22).

*5.22

TENSORS AND DIFFERENTIAL FORMS WITHOUT COORDINATES

-,<,'>

The elaborate formalism of tensors and differential forms is due to their expressions in different coordinate systems. It is natural to seek ways of describing the objects and operations in so far as possible without coordinates. An example is provided by the interpretation of the gradient of a function F in Section 2.14 as a vector whose direction is that in which F increases most rapidly and whose length equals that rate of increase. Throughout, all functions will be assumed differentiable as needed. We begin by considering vectors in space, considered here as bound vectors as in a vector field. With each vector v at a point P we can associate the operator v . V (Problem 12 following Section 3.6). When this operator is applied to a function f , the result is a scalar v . grad f . (If v is a unit vector, then this equals the directional derivative o f f at P in the direction of v, but we do not assume that v is a unit vector.) We write L for the differential operator v .V and verify that L has the two properties: .

+

+

I . L(CIf~ c 2 f d = CILfl c2Lf2 (linearity), 11. L( f , f2) = f 1 Lf2 f2L f l (product rule).

+

(In I, c , , c2 are arbitrary scalars.) We now de$ne the vectors at P to be the set of all linear operators L satisfying the rules I and I1 at P for all functions f l , f2 differentiable in a neighborhood of P. One shows that each such operator L must be representable as v . V for a unique v (so that there is a one-to-one correspondence between the old vectors v and operators L). The definition applies to n-dimensional space. The set of all vectors at a point P (by the new definition) is called the tangent space at P. The term is suggested by the fact that we can apply the same idea to vectors tangent to a curve, surface, or hypersurfaces at a point P (cf. the "directional derivative along a curve" in Section 2.14). The operators L satisfying I and I1 at P for all functions defined and differentiable

371

372

Advanced Calculus, Hfth Edition in a neighborhood of P are again representable as the operators v . grad (where, for the curve or surface or hypersurface S, v is tangent to S). The operators L at a point can be added and multiplied by scalars and form a vector space. The dimension of this vector space is n if we are considering vectors in n-dimensional space or a hypersurface of dimension n. We next observe that, in an arbitrary vector space of dimension n, we can define a linear function as a real-valued function o satisfying the rule: 111. w(au

+ bv) = aw(u) + bo(v)

for all vectors u, v in the space and all real scalars a , b. In 3-dimensional space v3 an example of such a function is vo . v for fixed vo; in fact, every linear function on v3can be so represented, and an analogous statement holds for Vn. Again the functions o satisfying I11 can be added and multiplied by scalars and themselves form a vector space (also of dimension n). This vector space is called the dual space of the vector space we started with and its elements w are referred to as covectors. We apply this idea to the tangent space at P introduced above. The space of covectors is called the cotangent space at P. Finally we consider a function assigning a real number to each choice of p vectors L , , . . . , L , in the tangent space at P and of q covectors o l , . . . , wq in the cotangent space at P and satisfying a rule of multilinearify: @ is to be linear in each variable. For example, for p = 2, q = 1,

We define each such function @ to be a tensor at P, covariant of order p and contravariant of order q . As usual we include as tensors with p = 0, q = 0 the scalar functions defined at P. One shows that, if coordinates are introduced with reference to appropriate bases in the tangent space and its dual, then @ becomes a real function of 2n real variables a ' , . . . , an,PI, . . . , B,: @

"

J'

= Ail ...i,,

'1

. ..

, pi, . . . 4

(summation convention), where the coefficients A::: define a tensor at P of the type indicated. Thus we have achieved (in outline) our goal: a definition of tensors which does not rely on coordinates. Of course the coordinates are in the background and do motivate the definition. We have emphasized tensors at a point. By allowing P to vary over an open region, we obtain vector fields, covector fields, and tensor fields as in Section 3.9. For the tensor fields 0 becomes a function of P, in addition to L . . . , wl , . . . , and one assumes appropriate differentiability with respect to P.

4

373

Chapter 5 Vector Integral Calculus In the discussion at the end of Section 5.21 differential forms are related to alternating covariant tensors. One can in fact define a differential form of order m as a multilinear function @ of tangent vectors L1, . . . , L, (depending differentiably on P), which is alternating: that is, @ = @ ( P ,L1, . . . , L,), where

and, in general, @ reverses sign when two of the Li are interchanged. (For m = 0 or 1 , the condition is satisfied automatically.) In a coordinate system, the tangent vectors can be represented in terms of components d x l , . . . , dxn and, as in (5.216) (Section 5.21),

For a full discussion, one is referred to the book of Boothby cited at the end of the chapter.

PROBLEMS

+

+-

1. 1n E~ let CY = x1x4(0-form), = x4dx1 - x1dx4, = x1dx2dx3 x2dx3dx1 x3dxl dx2.Calculate and simplify: a) b) B 2 - Y c) BY +yB d) y2 e) d a f ) d2B g) dy i. 2. Illustrate by examples of 1-forms and 2-forms in E~ that ar = arl and B = jJ1imply a B = CYI PI and CYB= a1B1. 3. Prove the rule CYB= (- l)m"'QBa of (5.207). [Hint: By the distributive laws (5.208) it suffices to prove it for ,A

+

+

a = Adxi1 . . . dxl"'l, jl = Bdxj'

.. . dxh,.

;3

If i k = j e for some k, l ,both sides are 0. Otherwise, as in Example 2 of Section 5.19, one hastoverifythatonecangofrom(iI, ..., i,,, j l , ..., jrn,)to(jl,...,jm,,il, ..., i,,) by m 1 m2 interchanges.] 4. a) Prove the rule (5.211). b) Prove therule(5.212)forml = l , m 2 = l , n = 4 . c) Prove the rule (5.213) for m = 1, n = 4. 5. For y as in Problem 1, apply the given change of variables to obtain a form jj in f l , . . . , f and simplify: a) x1 = f 1 + f 2 , ~2 = 22 + 2 ~ 3x3 , = f 3 + 3 ~ 4x4 , =4 ~+ 1 ~4 b) x' = (~112,~2 = ~ 1 ~x32 =, 5123, x4 = i l i 4 . 6. Under the (invertible) change of coordinates in E ~x1 : = f I , x 2 = z1 f 2 ,z3 = f l f 3 , a smooth path C: x' = xl(t),a i t 5 b and a smooth surface S: x1 = xi(u, v), i = 1,2,3, (u, v) in R,,, become and 3 respectively. Let XI denote x l ( x l ,x2,x3) with x', x2,x3 expressed in terms of f l , f 2 ,f 3 . a) Express the line integral JcXl dxl X2dx2 X3 dx3 as an integral over

+

+ z. dx2dx3+ X2 dx3dxl + X3 dxl dx2 as a surface

+

b) Express the surface integral JJsXl integral over 3.

+

374

Advanced Calculus, Fifth Edition

7. In Section 5.21 it is asserted that each m-form in En can be written uniquely in the form satisfies (5.217) and (5.218). With the help of the alternation (5.216) where operation d of Section 3.1 1 prove this a) form = 2 b) for arbitrary m 5 n. 8. Let Ai and Bij be alternating covariant tensors in an open region of E ~ Show . that the exterior product of A; dx' and B ~ dx' , d x j equals heijk dx' d x J d x k , where Cijkis the exterior product of the tensors Ai and Bij. 9. State Stokes's theorem (5.215) explicitly in terms of coordinates in E ~ . a) for line integrals ( m = 1) b) for surface integrals ( m = 2). 10. (Difficult). Let Aij be a covariant alternating tensor in an open region of En(n > 3) and let

Let an allowable change of coordinates (5.219) be made under which the Aij become Aij and Bkij become Bkij (an in (5.220)). Show that in the new coordinates d ( & A i jd.ti d ~ j=) $Bkijd i k d?' d f J .

Suggested References Boothby, William M., An Introduction to Differentiable Manifolds and Riemannian Geometry, 2nd ed. revised, New York: Academic Press, 2002. Brand, Louis, Vector and Tensor Analysis. New York: John Wiley & Sons, Inc., 1947. Buck, R. C., Advanced Calculus, 3rd ed. New York: McGraw-Hill, 1978. Courant, Richard J., Differential and Integral Calculus, trans. by E. J. McShane, Vol. 2. New York: Interscience, 1947. Flanders, Harley, Differential Forms, with Applications to the Physical Sciences. New York: Academic Press, 1963. Z Franklin, Philip, A Treatise on Advanced Calculus. New York: Dover, 1964. Gibbs, J. W., Vector Analysis. New Haven: Yale University Press, 1913. Goursat, ~ d o u a r dA, Course in Mathematical Analysis, trans. by E. R. Hendrick, Vol. 1. New York: Dover Publications, 1970. Hodge, W. V. D., The Theory and Applications of Harmonic Integrals, 2nd ed. Cambridge: Cambridge University Press, 1952. Kellogg, 0. D., Foundation of Potential Theory. Berlin: Springer, 1967. Lamb, H., Hydrodynamics, 6th ed. New York: Dover, 1945. Sternberg, Wolfgang J., and Turner L. Smith, The Theory of Potential and Spherical Harmonics. Toronto: The University of Toronto Press, 1946. '

Infinite Series

6.1 INTRODUCTION An infinite series is an indicated sum of the form: going on to infinitely many terms. Such series are familiar even in the simplest operations with numbers. Thus one writes:

this is the same as saying

Of course, we do not interpret this as an ordinary addition problem, which would take forever to carry out. Instead we say, for example, 113 equals 0.33333 to very good accuracy. We "round off" after a certain number of decimal places and use the resulting rational number

i.

as a sufficiently good approximation to the number The procedure just used applies to the general series a1 a2 + . . a, . . . . To evaluate it, we round off after k terms and replace the series by the finite sum

+

+ +

376

Advanced Calculus, Fifth Edition However, the rounding-off procedure must be justified; we must be sure that taking more than k terms would not significantly affect the result. For the series I+-.., 1+1 such a justification is impossible. For one term gives 1 as sum, two terms give 2, three terms give 3, and so on; rounding off is of no help here. This series is an example of a divergent series. o n t h e other hand, for the series 1 1 1 1 1+ -+-+-+...+-+... .-4 9 16 n2 it seems safe to round off. Thus one has as sums of the first k terms: +

.

.

a

+

The sums do not appear to change much, and one would hazard the guess that the sum of 50 terms would not differ from that of four terms by more than, say, $. While it will be seen below that such appearances can be misleading, in this case our instinct happens to be right. This series is an example of a convergent series. It is the purpose of the present chapter to systematize the procedure indicated and to formulate tests that enable one to decide when rounding off is meaningful (convergent case) or meaningless (divergent case). As the example of the decimal expansion of shows, the notion of infinite series lies right at the heart of the concept of the real number system. Accordingly, a complete theory of series would require a profound analysis of the real number system. It is not our purpose to cany this out here, so some of the rules will be justified in an intuitive manner. For a complete treatment the reader is referred to the books by Hardy and Knopp listed at the end of the chapter. Sections 2.23,4.10, and 4.1 1 cover real variable theory in more depth and provide proofs for some of the key results of this chapter. Because of the large number of theorems appearing in this chapter, the theorems will be numbered serially from 1 to 59 throughout the chapter.

4

If to each positive integer n there is assigned a number sn,then the numbers s, are said to form an infinite sequence. The numbers are thought of as arranged in order, according to subscript: Examples of such sequences are the following:

377

Chapter 6 Infinite Series '

These are formed by the rules:

r

.

At times it is convenient to number the members of the sequence starting with 0, with 2, or with some other integer. A sequence s, is said to converge to the numbers s or to have the limit s: lim s, = s

(6.4)

n+w

if to each number t > 0 a value N can be found such that

I S , - S ( < E forn > N.

(6.5)

;

This is illustrated in Fig. 6.l(a). If s, does not converge, it is said to diverge. The limit s is clearly unique. For if s' is a limit different from s , we let 6 = 1s - s 1 ) / 2Then . as in (6.5), Is, - sl < c for n > N , and similarly, Is, - s'l < c for n > N'. Therefore if No is the larger of N and N', for n > No we have both Is, -s I < t and Is, - s'l c t , so that

+ Sn - 8'1 5 I S

26 = I S - s I I = ( S - 8,

- s,I

+ IS,

- s'I < 6

+

€

=2t,

1

and that is impossible.

, B

A

a. Convergent

I

I I

,I

s,

I

I

I:

2 monotone increasing

I

I

Is1

s2

4

3

S,

I I

I

I s.3

B

-A

c. Convergent monotone decreasing

d. Divergent monotone increasing

I

III I

I

I

I

S2

I

I

I

I

I

Figure 6.1

S,

I

S3

s1S2

$4

$5

B

A

e. D~vergent bounded

4

I

111111 I

Qpes of sequences.

i I 111 I

I

+s

378

Advanced Calculus, Fifth Edition

By a similar reasoning, other limits in the calculus are unique (for example, limits of functions, the definite integral as a limit). One can regard a sequence s, as a function s(n) of the integer variable n. The limit definition (6.4), (6.5) is then formally the same as that for a function f ( x ) of the real variable x . The sequences (6.1) and (6.2) can be shown to converge:

while (6.3) diverges. . ~-dnriiorr ,,i . A sequences, is said to be monotone increasing if s l 5 s2 I . . . I S, 5 S,+I . . ., this is illustrated in Figs. 6.1 (b) and (d). Since the = sign is permitted, it is perhaps clearer to call these sequences monotone nondecreasing. Similarly, a sequence s, is called monotone decreasing (or monotone nonincreasing) if s, 2 s,+~for all n; this is shown in Fig. 6.l(c). A sequence s, is said to be bounded if there are two numbers A and B such that A ( s, 5 B for all n. All the sequences suggested in Fig. 6.1 are bounded except those of (d) and (f).

THEOREM 1 Every bounded monotone sequence converges. Thus let s, be monotone increasing as in Fig. 6.l(b). The figure suggests that since the numbers s, move to the right as n increases but are not allowed to pass B, they must pile up on a number s to the left of B or at B. One can determine the decimal expansion of s as follows: Let kl be the largest integer such that s, kl for n sufficiently large; in Fig. 6.l(b), kl = 4. Let k2 be the largest integer between 0 and 9 (inclusive) such that

for n sufficiently large; in Fig. 6.l(b), k2 = 3. Let k3 be the largest integer between 0 and 9 such that

for n sufficiently large. Proceeding thus, we obtain the complete decimal expansion of the limit s:

If s = 4 $, one would find

--

.

1

A careful perusal of the discussion just given shows that the number s has actually been precisely defined as a real number and that the difference s - s, can be made as small as desired by making n sufficiently large. Thus Theorem 1 is proved. (See Section 2.23 for a more detailed proof.)

Chapter 6 Infinite Series

379

If a sequence s, is monotone increasing, there are then only two possibilities: The sequence is bounded and has a limit, or the sequence is unbounded. In the latter case, for n sufficiently large the sequence becomes larger than any given number K: s, >

K

forn > N;

:

(6.6)

this we also describe by writing: lim s, = GO,

n+m

or "s, diverges to oo." This is illustrated by the sequence s, = n ; see Fig. 6.l(d). It can happen that lim s, = oo without s, being a monotone sequence; for example, S, = n (-1)". Similarly, one defines:

+

.

lim s, = -00,

%.

(6.8)

n-tco

if for every number K an N can be found such that Every monotone decreasing sequence is either bounded or divergent to -oo. THEOREM 2

Every unbounded sequence diverges.

.

-

ProoJ: Suppose on the contrary that the unbounded sequence s, converges to s ; then

for n > N. Thus all but a finite number of s, lie in the interval between s - 6 and s + E , and the sequence is necessarily bounded, contrary to assumption.

6.3 UPPERAND LOWERLIMITS It remains to consider sequences s,, that are bounded but not monotone. These can converge, as in Fig. 6.l(a), but can equally well diverge, as the example 1, - l , l , -1,. . . , ( - I ) ~ + ' , .. . illustrates. This example also suggests the way in which a general bounded sequence can diverge, namely, by oscillating between various limiting values. In this example the sequence oscillates between 1 and -1. The sequence oscillates with three limit values 1,0, -1; the oscillation suggests trigonometric functions. In fact, the general term of this sequence can be written thus: 4

= sin ( i n n ) .

In general the bounded sequence can wander back and forth in its interval, coming arbitrarily close to many (even all) values of the interval; it diverges not for lack of limiting values, but because it has too many, namely, more than one.

'

380

Advanced Calculus, Fifth Edition Of greatest interest for such bounded sequences are the largest and least limiting values, which are termed the upper and lower limits of the sequence s,. The upper limit is defined as follows: lim s, = k n+co

if, for every positive E ,

1

IS, - kl

<E for infinitely many values of n and if no number larger than k has this property. Similarly one defines the lower limit:

if, for every positive 6 , 4

Is,, - hl < 6 for infinitely many values of n and if no number less than h has this property. It is a theorem that every bounded sequence has an upper limit and a lower limit; this is proved in much the same way as Theorem 1.(The proof of Theorem C in Section 2.23 shows the existence of a finite upper limit so for a sequence that is bounded above.) EXAMPLES

-

lim (-1)" = 1,

n+w

n-tcc

lim (-1)"

= -1,

n+w

sin ( i n n ) =

1

lim sin ( i n n ) = -I1 A, -

n+w

-

lim (-lln(l

n+m

+ l/n) = 1,

lim

n+w

sinn = 1,

lim (-l)"(l -

n+w

+ l/n) = -1,

lim sinn = -1. -

n+w

These should be verified by graphing the sequences as in Fig. 6.1. The proofs of the relations are not difficult, with the exception of the last two, which are quite subtle. The definitions of upper and lower limits can be extended to unbounded sequences. One must then consider oo and -oo as possible "limiting values"; thus oo is a limiting value if for every number K, s, > K for infinitely many n; -oo is a limiting value if for every number K, s, < K for infinitely many n. The upper limit is then defined as the "largest limiting value" and the lower limit as the "smallest limiting value." EXAMPLES

-

lim (-1)"n = oo,

n+w

limn2sin2(fnn)=m,

n+w

b (-1)"n = -00,

4

n+cc

lirnn2sin2(knn)=~. -

n+w

9

iI

i

381

Chapter 6 Infinite Series Every sequence will then possess upper and lower limits. It should be noted that when lim s, = -oo,the upper and lower limits are both -oo. In general, the lower limit cannot exceed the upper limit. THEOREM 3 If the sequence s, converges to s, then

-

lim sn = !@ s, = s;

n+oc

n+cc

conversely, if upper and lower limits are equal and finite, the sequence converges. 1 The first part of the theorem follows from the definition of convergence; for the limit s satisfies all conditions for both upper and lower limits. If the upper and lower limits are both finite and equal to s, the sequence is necessarily bounded, for otherwise the upper limit would be +oo or the lower limit would be -00.If s, does not converge to s, then for some E there are infinitely many members of the sequence such that Is, - s I > E . One can then proceed as in the proof of Theorem 1 to produce a limiting value k other than s. This contradicts the fact that s is both the largest and . least limiting value.

We first point out connections between sequences and continuity of functions. 3ms

THEOREM 4 xo, then

Let y = f (x) be defined for a 5 x 5 b. If f (x) is continuous at i

t

for every sequence x, in the interval converging to xo. Similarly, iff (x, y) is defined in a domain D and is continuous at (xO,yo), then

for all sequences x, , y, such that (x, , yn)is in D and x, converges to xo, y, converges to Yo. We prove the statement for f (x), the proof for a function of two (or more) variables being similar. Given E > 0, a 8 can be found such that If (x)- f (xo)l < E for Ix -xo I < 8 ; this is simply the definition of continuity. One can then choose N so that Ix, -xol < 6 for n > N, sincex, converges toxo. Accordingly, If (x,,) - f (xo)l < 6 for n > N; that is, f (x,) converges to f (xo). We remark that there is a converse to Theorem 4: If f (x) has the property that f (x,,) converges to f (xo) for every sequence x, that converges to xo, then f (x) must be continuous at xo (Theorem G in Section 2.23).

. .

- I

Advanced Calculus, Fifth Edition THEOREM 5

If

then

provided, in the last case, that there is no division by 0.

Proof. The function f ( x , y) = x Theorem 4

+ y is continuous for all x and y. Hence by

lim f (x,, y,) = n+cc lim (x,

n+m

+ y,) = x + y .

The other rules follow in the same way, with f ( x , y ) chosen successively as x - y, x . y, x l y .

THEOREM 6 (Cauchy Criterion) If the sequence s, has the property that for every 6 > 0 an N can be found such that

IS,

-s,I

< E

forn > N and m > N,

(6.10)

then s, converges. Conversely, if s, converges, then for every r > 0 an N can be found such that (6.10) holds. Proof. If (6.10) holds, then the sequence s, is necessarily bounded. Indeed, if we choose a fixed 6 and N according to (6. lo), then for a fixed m > N,

for all n greater than N. Hence all but a finite number of the s, lie in this interval, so that the sequence must be bounded. If the sequence fails to converge, then by Theorem 3 the upper limit S and lower limit E must be unequal. Let S - 5 = a . Then

for infinitely many m and n. Hence Ism - s, I > a / 3 for infinitely many m and n, as Fig. 6.2 shows. If 6 is chosen as a / 3 , condition (6.10) cannot be satisfied for any N. This is a contradiction; hence S = E, and the sequence converges. Conversely, if sn converges to s , then, for a given r > 0 , N can be chosen so that Is, - sl < ! p for n > N. Hence form > N and n > N,

Is~-s~I=Is~-s+s-s~I~

IS~-SI+IS-S,(<

Accordingly, condition (6.10) is satisfied.

Figure 6.2

Cauchy condition.

f € + ! j ~ = ~ .

Chapter 6 Infinite Series > .

PROBLEMS

383 :q

1. Show that the following sequences converge and find their limits:

+ k),

d) n log (1 e) s , = 1 f o r n = 1 , 2 , 3 ,... 2. Determine the upper and lower limits of the sequences: a) cos nn, b) sin f nn, 3. Construct sequences having the properties stated:

t

a.

C)

.

n sin ~1 n n .

-

limn,oosn = +oo, m il , s, = +oo. Show that the Cauchy criterion is satisfied for the sequence s, = lln. Evaluate e to 2 decimal places from its definition as limit of the sequence (1 + lln)". Evaluate lim,,,xn and lim,,,xn. The number n can be defined as the limit, as n + oo,of the area of a regular polygon of 2" sides inscribed in a circle of radius 1. Show that the sequence is monotone and use it to evaluate n approximately. C)

4. 5. 6. 7.

I

An infinite series is an indicated sum: of the members of a sequence a,. The series can be abbreviated by the

n=l

C sign: sdF

notation is to be preferred, except for very simple series. and the Associated with each infinite series C a, is the sequence of partial sums S,:

(Note that the index j in the sum here is a dummy index, so that the result does not depend on j. Thus C;=,a, = C;=, a,. This is like the case of the integration variable in a definite integral.) Accordingly,

S1= a ! ,

S2=al+a2,

. . . , Sn = a l + . . - + a , .

DEFINITION The infinite series Czla, is convergent if the sequence of partial sums is convergent; the series is divergent if the sequence of partial sums is divergent. If the series is convergent and the sequence of partial sums S, converges to S, then S is called the sum of the series, and one writes

384

Advanced Calculus, Fifth Edition ""'l st-<

Thus by definition,

r*.

when the limit exists. If lim S, = +oo or lim S, = -00, the series a, is said to be properly divergenr. From Theorem 1 of Section 6.2, one concludes: THEOREM 7 If a, > 0 for n = 1 , 2 , . . . , then the series gent or properly divergent. For, since a, 0,

xz,

a, is either conver-

that is, the S, form a monotone increasing sequence. If this sequence is bounded, it converges by Theorem 1; if it is unbounded, then necessarily lim S, = ca,so that the series is properly divergent. EXAMPLE 1 The series

is properly divergent. For each term is at least equal to

i, and accordingly,

The sequence S, is monotone but unbounded, and lim S, = oo. It should be noted that the terms of this series form a convergent sequence; for

This does not imply convergence of the series. In general, one must carefully distinguish the notions of series, sequence of terms of a series, and sequence of partial sums of a series. The series is simply another way of describing the sequence of partial sums; the terms of the series describe the steps taken from each partial sum to the next. This is suggested in Fig. 6.3.

I

-02

I S,= a l

O

Figure 63

Series

I

a7-ra4-

I

I

I

S2

S,

S4

C a, versus sequence S,, of partial sums.

WX

385

Chapter 6 Infinite Series

EXAMPLE 2

,

The series

.

J

1

is convergent. Here the partial sums form the sequence

which converges, as before, to 1. Hence

From Theorem 5 of Section 6.4, one deduces a rule for addition and subtraction of convergent series, as well as a rule for multiplication by a constant:

xzl

a, and THEOREM 8 If tively and k is a constant, then

xzlb, are convergent with sums A and B respecif

For if one introduces the partial sums:

A, = a ,

+

a

*

.

+

Sn = (a1

+

B, = b , + . . . + b , ,

a,,

+ 61)

+ . a .

+ (a, + b , ) ,

then S, = A, B,. Since A, converges to A and B, to B, S, converges to A A similar reasoning holds for the series C ( a , - b,), C ka,.

+ B.

The Cauchy criterion (Theorem 6) can also be interpreted in terms of series: THEOREM 9 The series C z la, is convergent if and only if to each E > 0 an N can be found such that

la,+l + a , + 2 + . . . + a m I

< E

form > n > N.

(6.16)

This is a simple rewriting of condition (6.10). For, when m > n,

6.6 TESTSFOR CONVERGENCE AND DIVERGENCE

I

A topic of prime importance is the formulation of rules.or "tests" that permit one to determine whether a particular series converges or diverges. The problem is similar to and, in fact, closely related to that of improper integrals. This connection will be made clear later.

386

Advanced Calculus, Fifth Edition

A few preliminary remarks will be helpful. The convergence or divergence of a series is unaffected if one modifies afinite number of terms of the series; thus the first ten terms can be replaced by zeros without affecting the convergence, though it will of course affect the sum if the series converges. The terms of the series can all be multiplied by the same nonzero constant k without affecting convergence or divergence, for the partial sums S, converge or diverge according to whether kS, converges or diverges. In particular, all terms can be multiplied by - 1 without affecting convergence; however, other changes in sign can change the character of the series completely. The operations of grouping terms and of rearrangement of the terms of a series require considerable care. They will be discussed briefly later. THEOREM 10

(The nth Term Test) If it is not true that lim a, = 0,

n--f,

then

Czla,

ri

diverges.

Proof. If the series were to converge to S,then lirn a, = n+w lirn (S,+I - S,) = n+w lim

n+oo

- lirn S, = S - S = 0. n-03

Hence if a, does not converge to 0, the series cannot converge. It is to be emphasized that this test can be used only to prove divergence. If lirn,,, a, = 0, the series C a, may converge or diverge. It should also be noted that to show that a, does not converge to 0, one need not show that a, converges to a number different from 0, for the same conclusion is reached if one can show that the sequence a, diverges.

EXAMPLES 1. C,"O=,(-1)". Here a, = f 1, hence a, diverges and the series diverges. 2. C,"O=, n. lirn,,, a, = limn,, n = m. The series diverges. 3.

4.

Czl -. limn,, C,"O=, A. lirn,,, a,

3 The series diverges. a, = a.

= 0. The test proves nothing. The series is the harmonic series and will be shown to diverge. 5. Czl limn+, a, = 0. The test proves nothing. The series converges to 1, as shown in Section 6.5.

&.

If a series a, is such that absolutely convergent.

la, I is convergent, then the series

a, is called

THEOREM 11 (Theorem on Absolute Convergence) If Czl (a, ( converges, then C,"=, a, converges; that is, every absolutely convergent series is convergent. Proof. The theorem follows from the Cauchy criterion (Theorem 9 of Section 6.5). For

Chapter 6 Infinite Series If ELP=, la,l converges, then the s<m on the right is less than r when n 5-A? for suitable choice of N; hence the sum on the left is less than c for n > N, so that the series C a, converges. One can interpret this theorem as stating that introduction of minus signs before various terms of a positive term series tends to help convergence; if the original series diverges, introduction of enough minus signs may make the series converge; if the original series converges, introduction of the minus signs will make the series converge even more rapidly. .s

EXAMPLES

6.

I)" C z , -(-iR = 1 - f + f . . . . Since the series CzI$ converges, this series

converges. 7. Cz,(-I)". The series of absolute values is 1 1 . . . 1 . . . This series diverges, so that Theorem 11 gives no help. However the nth term fails to converge to 0, so that the series diverges.

+ +

+

+ +

+

= 1-1 1 - 41 . . . . The series of absolute values is the 2 3 harmonic series of Example 4. Although the harmonic series diverges, it will be seen that this series converges.

(-')" s. C,"=I =i-

A series C a, that converges but is not absolutely convergent is called conditionally convergent. Such a series converges because the minus signs have been properly introduced. An example is the series of Example 8. THEOREM 12 (Comparison Test for Convergence) If la, 1 5 b, for n = 1, 2, . . . and bn converges, then Czlan is absolutely convergent.

cz,

Proof. By Theorem 7 the series C la, I is either convergent or properly divergent. If it were properly divergent, then

since b, 2 la, 1, one would then have

so that b, would diverge, contrary to assumption. Hence C la, I converges, and a, is absolutely convergent. EXAMPLE 9

CE, &.Since

"

1

and the series C 112" converges, the given series converges. THEOREM 13 (Comparison Test for Divergence) 1,2, . . . and Czlbn diverges, then Czla, diverges.

If a, I b, 2 0 for n =

387

388

Advanced Calculus, Fifth Edition

Proof. If C a, were to converge, then 1b, = 12. Hence C a, diverges.

(b,1 would converge by Theorem

4

•

It should be emphasized that the test of Theorem 13 applies only to positive term series.

xz,9.

EXAMPLE 10 It is apparent that the terms are close to those of the divergent harmonic series CE=, However, the inequality

i.

is false since it is equivalent to the statement n2 - n > n2. On the other hand,

.I 4

n-1 1 >n2 2n

. 'is:

for n > 2, since this is equivalent to the inequality: 2n2 - 2n > n2 or to the inequality n > 2. The constant factor and the failure of the condition for n < 3 have no effect on convergence or divergence. Hence Theorem 13 applies, and one concludes that the series diverges. a It should be stressed that one can conclude nothing about the series C a, from the inequality la, I I Ib, I, where C b, diverges; nor can one conclude anything from the inequality a, 2 b, > 0, when C b, converges. THEOREM 14

(Integral Test)

Let y = f (x) satisfy the following conditions:

a) f (x) is defined and continuous for c 5 x < oo; b) f (x) decreases as x increases and lim,,, f (x) = 0; C)

I

f(n) =an.

Then the series Czla, converges or diverges according to whether the improper integral 1: f (x) dx converges or diverges. Proof. Let us suppose the improper integral converges. Assumptions (b) and (c) imply that a, > 0 for n sufficiently large. Hence by Theorem 7 of Section 6.5 the series C a, must either converge or properly diverge. Let the integer m be chosen so that m > c. Then, since f (x) is decreasing,

Ln+l

f (x) dx

> f (n + 1) = a,+l

+

as illustrated in Fig. 6.4. Hence 0 < a,+l f (x) d x . It is thus impossible for the series series must be convergent.

for n

> m,

-'.+ am+, 5

"'+P lrn f(x)dx

I

a, to diverge properly, and the

•

The proof in the case of divergence is left as an exercise (Problem 10 following Section 6.7).

Chapter 6 Infinite Series i

Figure 6.4

,

Proof of integral test for convergence.

THEOREM 15 The harmonic series of order p:

converges for p > 1 and diverges for p 5 1.

I

'

>t

'zff

h o o j The nth term fails to converge to o when p $%'gat thb'snies surely diverges when p ( 0. For p > 0 the integral text can be applied, with f (x) = 1/xP. Now for p # 1,

This limit exists and has the value 1/(p - 1) for p > 1. For p < 1 the integral diverges, and for p = 1,

Ilm :

- d x = lim logb = oo, b-tm

so that again there is divergence. The theorem is now established. When p = 1, one obtains the divergent series

commonly called the harmonic series; the connection with harmony (that is, music) will be pointed out in connection with the Fourier series in the next chapter. For general (even complex) values of p the series of Theorem 15 defines a function {(p), the zeta-function of Riemann.

THEOREM 16 The geometric series

converges for - 1 -c r < 1:

and diverges for Ir 1

> 1.

r

390

Advanced Calculus, Fifth Edition

Prooj

The algebraic identity

I

+ r + . . - + rn-I)

1 - rn = (1 - r)(l

leads to the rule for geometric progressions:

Therefore for the geometric series,

1 -rn

S,, = a -

1-r

,fn4 B ZSPX:~IR~

T

+r

for r # 1. For - 1 c r < 1, rn converges to 0;hence S, converges to a/(l - r ) . For Irlz 1 the nth term of the series does not converge to 0, so that the series diverges. 7--

THEOREM 17

-

m,

-T!-

(Ratio Test) If a, # 0 for n = l , 2 , .. . and

then bO

a, is absolutely convergent,

if L < 1, n=l

if L = 1 ,

the test fails,

2

a, is divergent.

+

+

Prooj Let us assume first that L < 1. Let r = L (1 - L)/2 = (1 L)/2, so that L < r < 1 . Let b, = la,+l / a , 1. B y assumption, b, converges to L , so that 0 < b, < r for n > N and appropriate choice of N. Now one can write

Since 0 < b, < r for n > N, each term of the series in parentheses is less than the corresponding term of the geometric series

l a ~ +l l( 1

+ r + r 2 + r3 + . .

e).

This series converges, by Theorem 16, since r < 1. Hence by Theorem 12 the la, 1 converges, so that CEl '=,a, 1 converges and C a, is absolutely series CZN+, convergent. .

+

3

Chapter 6 Infinite Series

n

If L > 1, then la,+l /a, 1 2 1 for n sufficiently large, so that la, 1 Ila,+l 1 5 . . . . Since the terms are increasing in absolute value (and none is 0), it is impossible for a, to converge to 0. Accordingly, the series diverges. For L = 1 the series can converge or diverge. This is illustrated by the harmonic series of order p, for lim

n-tm

P

np

( n + l)P

= lim n-+w

(5) =I n+1

The limit of the ratio is 1; however, the series converges when p > 1 and diverges when p 5 1 by Theorem 15. THEOREM 18

(Alternating Series Test) An alternating series m

a , -a2+an -a4

+ . a .

= C(-l)"+'a., n=l

a, > 0.

converges if the following two conditions are satisfied:

a) its terms are decreasing in absolute value: a,+l ( a ,

f o r n = l , 2 ,...,

Proof. L e t S , = a l - a 2 + a 3 - a 4 + . . . f a n. T h e n S 1 = a 1 , S 2 = a l - a 2 < S 1 , S3 = S2 + a3 > S2, S3 = S1 - (a2- a 3 ) < S l , SO that S2 < S3 < S1.Reasoning in this way, we conclude that as shown in Fig. 6.5. Thus the odd partial sums form a bounded monotone decreasing sequence, and the even partial sums form a bounded monotone increasing sequence. By Theorem 1 of Section 6.2, both sequences converge: lim S2n+l= S*,

n-+m

lim S2, = S**.

n+m

Now lirn a2,+1 = lirn (S2,,+1- S2") = S*

n-rm

n-+m

- S**.

By assumption (b) this limit is 0, so that S* = S**. It follows that the series converges to S*.

Figure 6.5

Alternating series.

391

I 392

Advanced Calculus, Fifth Edition THEOREM 19

(Root Test) Let a series

xz,

a, be given and let

if R < 1, the series is absolutely convergent; if R > 1, the series diverges; if R = 1, the test fails.

rr.4

:
-

' r ~i i p i !

6

rip'

Proof. If R < 1, then, as in the proof for the ratio test, one can choose r and N such that r < 1 and forn > N.

r Hence

la,I < r n forn > N, and the series C la, 1 converges by comparison with the geometric series C r n . If R > 1, > 1 for n sufficiently large, so that la, 1 > 1, and the nth term cannot converge to zero. The failure for R = 1 is again shown by the harmonic series; thus fora, = l/nP, lim

n+m

= lim

1

1

- - nlim +w

n-+w n ~ / n

e(~/n)logn

1

=-=I, e0

since ( l l n ) log n converges to 0.These series converge for p > 1 and diverge for p I1; thus the root test can yield no information when R = 1.

6.7

EXAMPLES OF APPLICATIONS OF TESTS FOR CONVERGENCE AND DIVERGENCE

The results of the preceding section provide the following tests:

a) nth term testfor divergence (Theorem 10); b) comparison test for convergence (Theorem 12); c) comparison test for divergence (Theorem 13); d) integral test (Theorem 14); e) ratio test (Theorem 17); f) alternating series test (Theorem 18); g) root test (Theorem 19).

In addition, the theorem on absolute convergence relates the convergence of a series to that of the series of absolute values of the terms. It is simplest to remember this as a basic principle and then to regard the tests (b), (c), (d), (e), and (g) as tests for positive term series. The general theorems on sequences and series in Sections 6.2 to 6.5 can also be of aid in testing convergence. In particular, one can in certain cases use the definition

Chapter 6 Infinite Series of convergence directly; that is, one can show that S, converges to a definite number S. This was done earlier for the geometric series (Theorem 16). On occasion the Cauchy criterion (Theorem 9) can also be used. Theorem 8 suggests a method of proving convergence by representing the given series as the "sum" of two convergent series; this idea can also be used to prove divergence, as shown in Example 12. On the basis of these remarks we enlarge the preceding list to include the tests:

h) partial sum test; i) Cauchy criterion (Theorem 9); j) addition of series (Theorem 8). In general, the problem of deciding convergence or divergence of a given series can require great ingenuity and resourcefulness. A vast number of particular series and classes of series have been studied; it is therefore important to be able to use the literature on the subject. The book by Knopp cited at the end of this chapter is suggested as a starting point for this. Some additional information is given in Section 6.8 and in Chapters 7 and 8, but this can only be regarded as a sampling of a huge field. It is recommended that the nth term test (Theorem 10) be used first; if this yields no information, other tests can then be tried.

xz,

+4

= 1. - . . The series of absolute values is EXAMPLE 1 the harmonic series C which diverges by Theorem 15. The series is hence not absolutely convergent, so that tests (b), (d), (e) are of no use. Since the terms are of variable sign, (c) is of no use. The nth term does approach 0 as n becomes infinite, so that (a) yields no information. The alternating series test (f) is applicable, since 1 > > . . ., and the nth term approaches 0.Hence the series converges. Since it is not absolutely convergent, it is conditionally convergent.

i,

It can be shown (see Problem 1 following Section 6.16) that the sum of this series is log 2 = 0.693 15 . . . . There exist methods of applying (h), that is, of showing that the nth partial sum does approach log 2 as limit; however, these are too involved to describe here. The Cauchy criterion can also be applied. For one can show, without , difficulty, that

As a matter of fact, the Cauchy criterion could also be used to prove Theorem 18 on alternating series; it will always apply when Theorem 18 applies (cf. Problem 1I). The operations of addition and subtraction, at least in their simplest form, do not help here. 7 i

xEl(-I)"+' q. This at first suggests the alternating series test. + l)/n converges to 1, the nth term does not converge to 0.The

EXAMPLE 2 However, since (n series diverges.

zzl

EXAMPLE 3 ,n2;::+2. The nth term converges to zero, so that (a) is of no help. For large values of n the general term is approximately n/3n2 = 1/3n,

393

394

Advanced Calculus, Fifth Edition since the terms of lower degree in the numerator and denominator become small in comparison to those of highest degree. This suggests comparing with x ( l / 3 n ) for divergence. The inequality

is not correct. However, the inequality

.I

is correct for n > 2. Since C ( l / n ) diverges, C(1/4n) diverges, and the given series diverges. •

A.

The terms are smaller than those of the harmonic EXAMPLE 4 Zz, series, so that one might hope for convergence. However,

since all the conditions of the integral test are satisfied, the series diverges.

EXAMPLE 5

Em,%.Here there is no doubt concerning divergence. for logn 1 -> - (n L 3).

n n EXAMPLE 6 This is like Example 5, but the higher power of n makes a considerable difference. The function logn grows very slowly as n increases; in fact.

EzD_, y.

log n lim -= 0 for p > 0, as a consideration of indeterminate forms shows. Hence the inequality 1 1 -logn = - . - logn 7 n2 nTI n73
EXAMPLE 7

xz,$.

Here the ratio test applies:

2"+l n! 2 lim -. - - lim -- 0, n-rn(n l)! 2" n - r r n n + 1

? -3 r c r

+

since

--n!

-

1.2.3...n

( n + I)! 1 . 2 - 3 . - . n ( n + 1) Hence L = 0, and the series converges.

-- 1

n + 1' a

Chapter 6 Infinite Series

EXAMPLE 8 lim

n+"

zz,5.Again the ratio test applies: +

( n I)"+'n ! = lim n + l . ( 7 n) n + l ( n + l ) ! nn n+mn 1

+

= lim ( 1

+

n+w

:In.

The limit on the right is e; since e > 1 , the series diverges. One can in fact conclude from this result that . nn lim - = oo; n-+m n ! that is, the nth term approaches oo as n becomes infinite.

EXAMPLE 9

2" CEI1.3.5...(2,,+1).

n-+m

The ratio test applies: 1. 3 . 5 . - . ( 2 n + 1)

2n+1 lim

1 . 3 . 5 . . (2n + 1)(2n

2-.,

+ 3)

2"

= lim n-+w

2 - 0. 2n

+3

The series converges.

EXAMPLE 10 partial sum:

C z ,log *. Here the simplest procedure is to consider the 1 Sn = log 2

n + log -23 + . . - + log n+l

Thus lim Sn = -00, and the series diverges.

EXAMPLE 11 term is

ZEl

s.

Here the principle of addition applies, for the general

s

i

r

n2 + 2" -2"n2

1 + -.1 -2"

n2

+ $k

Since C ( 1 / 2 " )and C ( l / n 2 )converge, the given series converges.

h).

Here the addition principle applies in reverse. If EXAMPLE 12 C:, ($ this series converges, then the sum

would also have to converge. Hence the series diverges. In general, the "sum" of a convergent series and a divergent series must be divergent. However, the "sum" of two divergent series can be convergent; this is illustrated by the divergent series: :

395

Advanced Calculus, Fifth Edition which, when added, become the convergent series

xz2(,oiny. m=-1 n

Here the root test is easily applicable:

EXAMPLE 13

log '

so that the root approaches 0 as limit, and the series converges.

Em2(*).

EXAMPLE 14

. Again the root test proves convergence: lim

n+m

= lim

n-*m

n

- 0. 1 + n2

PROBLEMS 1. Prove divergence by the nth term test:

a)

CEl sin (q)

b)

CE, 2"7 I

2. Prove convergence by the comparison test: a)

00

Cn=2

1

b)

CEl $p

b)

CZ2

-w;

3. Prove divergence by the comparison test:

a)

CZl n 2 -n +3n5 - 5

1

4. Prove convergence by the integral test:

8)

00

En=,

1

c -c 00

b) C n = 2

-.-

1

5. Prove divergence by the integral test: 6. Determine convergence or divergence by the ratio test: a)

CZl 9

b)

7. Prove convergence by the alternating series test: b) a)

CE, g

CEl zn + 1 tr\

I'i 7 bn:

CE2(- 1)"n log n

8. Prove convergence by the root test: m 1 En=, ;;ii

(A)

b) CEI n2 9. Prove convergence by showing that the nth partial sum converges: a)

nfl n CZI ( n +2)(n1 + 1) = CZl (m - m) m 1-nb) C=l - CZ1(* - $1

a)

10. Prove the validity of the integral test (Theorem 14) for divergence. 11. Prove the validity of the alternating series test (Theorem 18) by applying the Cauchy criterion (Theorem 9).

Chapter 6 Infinite Series 12. Determine convergence or divergence: oo

n+4

a) C , = I 2n3 - 1

b,

C z 3nl -r5

.

c ) , C zen m

13. (Ratio form of comparison test) Let a. > 0 and b, > 0 for n = 1,2, . . . and let the sequence a,/b, have limit k , possibly infinite. a) Show that if 0 < k < ao,then the series C a, and b, either both converge or both diverge. b, implies convergence of 2 a,, but b) Show that if k = 0,then convergence of C b, may diverge while a, converges. c ) Discuss the case k = oo. [Hint: Use (b), reversing the roles of a, and b,.] 14. Apply the results of Problem 13 to test for convergence of the given series C a,.

E l 1 C ) C z l sin ; ;

a)

+

, (take bn = l / n )

b)

Eml

ni;:i25(take

d ) C g l ( l - cos

b, = 1/n2)

i)

RATIOTESTAND ROOTTEST *6.8 EXTENDED It may happen that the test ratio ( a , + , / a , I has no limit as n becomes infinite. The behavior of this ratio may still give information as to the convergence or divergence of the series.

THEOREM 20 If a, # 0 for n = 1,2, . . . and there exist a number r such that 0 < r < 1 and an integer N such that

then the series C z lan is absolutely convergent. If, on the other hand,

for some integer N, then the series diverges.

•

The proof of Theorem 20 is the same as that of Theorem 17; for in that proof the existence of the limit L is not actually used. In the case of convergence, one uses only the existence of the number r such that

for n > N; in the case of divergence, one uses only the fact that the terms are increasing in absolute value and cannot approach 0.

397

398

Advanced Calculus, Fifth Edition EXAMPLE

In the series

the odd terms and even terms each form a geometric series with ratio of successive terms are

$.The ratios

5

in general the ratio of an even term to the preceding one is and the ratio of an odd term to the preceding one is The ratios approach no definite limit but are always less than or equal to r = $. Accordingly, the series converges.

i.

It is of interest to note that part of Theorem 20 can be stated in terms of upper limits (Section 6.3): i

THEOREM 20(a)

If an # 0 for n = 1,2, . . . and

then the series C a, converges absolutely.

+

For if the upper limit is k < 1, then the ratio must remain below k 6, for n sufficiently large, for each given E > 0. One can choose 6 so that k 6 = r < 1, di. and Theorem 20 applies. It should be remarked that no mclusions can be drawn from either of the relations: /

+

- Qn+1 1

= l . ;.tl :> t

The condition

'

I-I

an+1 >1 an implies divergence, but this is not as precise as Theorem 20. The root test (Theorem 19) can be extended in the same way as the ratio test:

lim -

n-m

THEOREM 21 that w

If there exist a number r such that 0 < r < 1 and an integer N such

r

forn > N,

then the series C,dO_, a, converges absolutely. If, on the other hand, for infinitely many values of n, then the series diverges. The proof in the case of convergence is the same as that of Theorem 19. If 5 1 for infinitely many values of n, then la, I 2 1 for infinitely many values of n, so that the nth term cannot converge to 0, and the series diverges.

Chapter 6 Infinite Series The theorem has a counterpart in terms of upper limits: THEOREM 21(a)

then the series

If

EL,a, is absolutely convergent. If lim

,-too

then the series

rn > 1,

xzla, diverges.

•

"1

No conclusion can be drawn from the relation

as the harmonic series show. Again the previous proof can be repeated.

'

5-

WITH SERIES rn ESTIMATE OF ERROR *6.9 COMPUTATION

Up to this point we have considered only the question of convergence or divergence of infinite series. If a series is known to be convergent,one still has the problem of evaluating the sum. In principle, the sum can always be found to desired accuracy by adding up enough terms of the series, that is, by computing the partial sum Sn for n sufficiently large. However, "sufficiently large" can mean very different things for different series. In order to make the procedure precise, one must have information, for each particular series, as to how large n must be to ensure the accuracy desired. This amounts to saying that one must determine, for each series, a function N ( E )such that

in other words, the first N terms are sufficient to give the desired sum S, with an error less than E . One can phrase this in another manner. One can write

i

S=S,+R,,

where R, is the "remainder." One then seeks N ( 6 ) such that

I R, 1 < r

for n ? N ( r ) .

It will be seen that for certain convergent series, one can find a useful explicit upper estimate for R, ; more precisely, one can find a sequence Tn converging to 0 for which

If the sequence T, is monotone decreasing, then we can choose N ( c ) as the smallest integer n for which Tn < E . For, if N ( E )is SO chosen and n 2 N ( E ) ,then

399

400

Advanced Calculus, Fifth Edition It will now be shown how such a monotone decreasing sequence Tncan be found when the series converges by one of the following tests: comparison test, integral test, ratio test, root test, alternating series test.

If la, I

THEOREM 22

( bn for

n 2 n l and

C:,

bn converges, then

for n 2 n l ; the sequence Tn is monotone decreasing and converges to 0.

, t i

ProoJ: By definition

+ - - .= lim

R. = a n + l + . . - + a n + ,

x

"+P

a,. P+oom=n+l

Since, for n 2 n 1, lan+l

"52

+ . - + an+,, I 5 Ian+~1 + . . + lan+pl

,.

e,

m=n+l

one concludes that I R ~ I =Iim P+w

1

,IJ

"+P

tninq

x 00

aml 5 m=n+l

bm = T". m=n+~

Since Tn is the remainder, after n terms, of a convergent series of positive terms, it is necessarily a monotone sequence converging to 0. The theorem can be stated as follows: If a series converges by the comparison test, then the remainder is in absolute value at most equal to that of the comparison series. rrn can be compared with the geometric series EXAMPLE 1 The series C ( 1 / 2 " ) .Hence the remainder after five terms is at most

zz,&

+

+

2"-2 . . . = 2-". l'hgkondition T, < r leads to the In general, Tn = 2-"-I inequality 2" > l / r , so that n > -(log E / log 2); we can thus choose N ( E )as the smallest integer n satisfying this inequality. "

xzl

an converges by the integral test of Theorem 14, THEOREM 23 If the series with the function f ( x ) decreasing for x > c, then

I R ~
SW

'

.%*>t+Y

"E

f ( x ) d X = Tn

n

for n 2 c ; the sequence Tn is monotone decreasing and converges to 0.

tf

Chapter 6 Infinite Series

401

Proof. This theorem follows from Theorem 22. For one can interpret the improper integral as a series:

one then has

As in the proof of Theorem 14,

so that

I

EXAMPLE 2

For the harmonic series, with order p > 1, one finds

This result can now be used, by Theorem 22, for any series whose convergence is established by comparison with a harmonic series of order p. If, for example, p = 6, then Tn = O.2n-'; it follows that we can choose N ( t ) as the smallest integer n such that n5 > 0.26-I. Accordingly, to evaluate the harmonic series of order 6 with an error less than 0.0001, five terms are sufficient:

THEOREM 24

If

for n > nl, so that the series C a, converges by the ratio test, then

the sequence Tn is monotone decreasing and convetges to 0. If

then r will be at least equal to L. If

402

Advanced Calculus. Fifth Edition for n 2 nl,then

lan2+11 = T,*, n I nl; Ian+]l - lan+2l the sequence T i is monotone decreasing and converges to 0.

(6.19)

lRnl i

ProoJ: The first statement follows from Theorem 22 and the fact that the ratio test is a comparison of the given series with a geometric series. Thus one has, as in the proof of Theorem 17,

The second statement emphasizes the fact that if the test ratio converges to L, then the ratio cannot remain less than or equal to a number r less than L. Thus r 2 L. One can use r = L only when

for n > N, so that the limit L is approached from below. ...$ The third statement concerns the case in which the test ratios are steadily decreasing and hence approaching a limit L. One cannot use L in such a case; however, I for n 2 n l , so that under the assumptions made, one can use r = lan+2/an+l

This gives formula (6.19).

EXAMPLE 3

xz,,s. The test ratio is found to be

This converges to example,

EXAMPLE 4

.ifir

i but is always less than i. Hence we can use r = i and, for

C,"=I "

Again the limit is decreasing since

Here the test ratio is

1,but the ratio is always greater than f . The successive ratios are .

-

Chapter 6 Infinite Series as algebraic manipulation show3. Hence inequality (6.19) applies (for n 2 I), and one finds, for example,

THEOREM 25

If m

9

<

1

for n > n l , so that the series 1a, converges by the root test, then

I&I

5

- T,, n > n l . 2 1-r

lim ,+a,

m=~< 1,

then r will be at least equal to R. If 1 > la,+l 11/("+1)2 la,+211/(n+2)

74

$39.

for n 2 n l , then

The sequences T, and T,* are monotone decreasing and converge to 0. The proofs parallel those for Theorem 24, since again the test is based on comparison with a geometric series. EXAMPLE 5

Ez2&.

Here 1 m=log n

'

This is decreasing and less than 1 for n = 3,4, . . . . Thus, for example, 1 1 R5 = -+-+... (log 6)6 (log 7)' THEOREM 26

1

I

,

GzF =0.06 ....

l-log6

If the series

converges by the alternating series test, then .,*

-

0 < lRnl < a,+, = T,.

Hence N ( r ) can be chosen as the smallest integer such that a,+l < E .

403

I

404

Advanced Calculus, Fifth Edition The theorem can be stated in words as follows: When a series converges by the alternating series test, the error made in stopping at n terms is in absolute value less than the first term neglected. Proof of Theorem 26. As was pointed out in the proof of the alternating series test (Theorem 18), one has It follows that the sum S lies between each two successive partial sums (one odd, one even); S2n<S<S2n+lr

S,<S<S,-f-

t

Hence

- S2, < S2,,+, - Sh = a,+l, = S - S2,,-1 > SZ,,- S2,,-1 = -a2,,;

0 < Rz,, = S 0 > R2,-1 that is,

0 < R, < ab+l,

-a,

.~sI fa 3.1

< R2,,-~< 0.

This shows that for every n, 0 < IRnI < an+l, but actually tells more: R, is positive if n is even; R, is negative if n is odd. (-lyl+l

1 EXAMPLE 6 C z l 7. Here the partial sums are S1 = 1, S2 = 1 - 21 = 2, 5 1 1 S 3 = ,, S4= .... The preceding theorem asserts that lR11 < 2 , lRzl < 3 , IR3I < . . . or, more precisely, that < Rl < 0,O < R2 < < R3 < 0. Thus in particular, if three terms are used, the sum is between and - $ = In order to compute the sum with an error of less than 0.01, one would need 100 terms, for the lOlst term, 11101, is the first one less than 0.01.

i,

&,

3, -a 2 2

-4

PROBLEMS

&.

.*

1. Determine how many terms are sufficient to compute the sum with given allowed error E and find the sum to this accuracy: a)

C z l 21 , E = 1

~ik

O0 c=0.5 i) Cn=2 nqlogn

b)

zz1(+,

f

= 0.1

j ) C , " = l m2", c = O . l

+

--

-

.

2. Let C a, be the geometric series 1 r + r 2 + . . . a) Determine how many terms are needed to compute the sum with error less than 0.01 1 when r = Z, r = 0.9, r = 0.99.

405

Chapter 6 Infinite Series b) Show that for any positive c,

1% 1 < 6 when

and that no smaller value of n will suffice. c) Show that as r approaches 1, the number of terms needed to compute the sum with error less than a fixed t- becomes infinite. 3. Show that for p > 0 the sum of the series 1 - 1/2p + 1/3P - . . . is positive.

It has been pointed out (Theorem 8) that convergent series can be added and subtracted term by term:

We also know that a convergent series can be multiplied by a constant:

.cms~ 3x6

Three other operations will be considered in the present section-multiplication, grouping, and rearrangement. We consider first the operation of grouping, that is, of inserting parentheses in a series a,. For example, the series C an might be replaced by the series ., that is, by C bn, where bn = ah-1 a h . (a1 a21 (~2~-1

+ + +

+

+

+

xzO=,

THEOREM 27 If the series a n is convergent, then insertion of parentheses a,. If an is yields a new convergent series having the same sum as properly divergent, insertion of parentheses yields a properly divergent series.

zz,

z:,

Proof. The effect of insertion of parentheses is to cause one to skip certain partial sums. Thus the partial sums of the series are the partial sums S2, Sq, S6, . . . , S2,, . . . of the series C a,. If S. converges to S, then the new sequence obtained by skipping must also converge to S; if lim Sn = +m, then the sequence obtained by skipping must also diverge to +m. Thus the theorem follows. Since a positive term series is either convergent or properly divergent, one can insert parentheses freely without affecting convergence or sum in such a series. This can be used to prove divergence of the harmonic series, for one has

'

406

Advanced Calculus, Fifth Edition

i,

each set of parentheses contributes at least so that the general term of the grouped series does not converge to zero. For a divergent series of variable signs,-insertionof parentheses can on occasion produce a convergent series. Thus the series C(- l)"+' becomes with the parentheses inserted as shown. Therefore in testing a series of variable signs for convergence, parentheses should not be inserted. A series Cz==, b, is said to be a rearrangement of a series C z la, if there exists a one-to-one correspondence between the indices n and m such that a, = b, for corresponding indices. Thus the series

are rearrangements of each other. Also

is a rearrangement of the harmonic series. THEOREM 28 If C z ,a, is absolutely convergent and C:=, b, is a rearrangement of C:, a,, then C;==, b, is absolutely convergent and has the same sum as 00 Cn=lan.

Proof.

One has clearly

since every b, equals an a, for appropriate n and no two m 's correspond to the same n. lbm1 converges, so that C b, is absolutely convergent. Let S, Hence the series I,"==, and S be the nth partial sum and sum for C a, and let Sh and S' be the corresponding quantities for C b,. For given positive E , let N be chosen so large that 1 1 and (a,+l( . . la,+,l < - E 2 2 for n > N and p 2 1; such an N can be found, since C a, converges to S and since the Cauchy criterion holds for C la, I . For m sufficiently large, Sk will be a sum of terms including all of al . . . a, and perhaps more:

IS, - SI <

-E

+ +

where k l , . . . ,k, are all larger than n. Let ko = n Then

+po be the largest of these indices.

Figure 6.6

407

Chapter 6 Infinite Series

at.

Cauchy product of series. 7

Now

y

:

1

It follows that Sk converges to S, so that S' = S as asserted. The preceding theorem is of importance in connection with the multiplication of series. If one multiplies two series as in algebra:

one obtains a collection of terms of form a,bm without any special order. It will be seen that when C a, and C bm are absolutely convergent, the products a,b, can be arranged in order to form an absolutely convergent infinite series. It follows from Theorem 28 that the same result is obtained no matter how the products are arranged and that the same sum is always obtained. Of the many possible arrangements the following one, known as the Cauchy product, is most commonly used:

1

This is illustrated in Fig. 6.6. If the series a, and C b, are indexed beginning with n = 0 and m = 0, the Cauchy product of the series appears as

This series is suggested by the multiplication of two power series (to be considered later):

the terms of same degree in x being collected together.

,:x

a, and C,"==, bmare absolutely convergent, then THEOREM 29 If the series Ci, the products a,bm can be arranged to form an absolutely convergent series and

xzl

408

Advanced Calculus, Fifth Edition

Figure 6.7

Prooj:

Proof of Theorem 29.

$.a .

We choose the series C ci as the series

as suggested in Fig. 6.7. From this definition, one has in particular

L

-

and in general cl

+ (c2 + c3 + c4) + - - + (. .. + c,2) = (a1 + . . + a,)(bl + - . - + b,).

Similarly,

~ l +I c ~+) . . .+ ( - .. + Ic,zl) = (la11 + . + la,I)(lbl I + . . - + lb,O. Since the series C la,( and C Ib, I converge, the equation just written shows that Icl I + (lczl

+l

ni +

the left-hand side approaches a limit as n += oo; dropping parentheses, on the bacn is absolutely sis of Theorem 27, we conclude that 1Ic, 1 converges, so that 1 convergent. The preceding equation then shows that

the insertion of parentheses being justified by Theorem 27. The theorem is thus proved. a On the basis of Theorem 28, one can now use any arrangement desired for the c's, in particular that of the Cauchy product. One can also group the terms as suggested by the preceding power series multiplication:

The insertion of parentheses is justified by Theorem 27. If the series C a, and C b, converge but are not both absolutely convergent, one can nevertheless form the Cauchy product of the series as C c,, where c, = sob, a1b,-1 . . . a,-l bl a,bo. It can be shown that if the series C c, converges, then always

+

+ +

+

409

Chapter 6 Infinite Series Furthermore, if one of the two series a n , bn converges absolutely, then en must converge. For proofs, refer to Chapter X of the book by Knopp listed at the end of the chapter.

PROBLEMS 1. Assuming the relations (proved in the next chapter): 1 Jr4 O0 1 Jr2 n=l n=l

O0

1

Jr6

C2=al Ca=s* C2==. n=l

evaluate the series: 6 a) CZO=1;;2

b)

2n2 - 3

n2+1 CEl ;;b

C) CZl -;;;r

2. Verify the following relations: m 1 . CEl 3 = Cn=2 (n - *)T * b) Z(", [f (n + 1) - f (n)] = limn,m

a)

f (n) - f (1). if the limit exists; C) z m 2 [ f ( n + 1)- f ( n - I)] = limn,,[f(n)+f(n+ 1)l- f(1)- f(2).iffielimit exists. 3. Use f (n) = 1/n and f (n) = 1/n2 in 2(b) and 2(c) to prove + 2 =1 b) CZ1 i q1~ r=)1 a)

CE1 ,, (,

+ I,

1

;=. ~- - r ;c); 2EC 4. From the relation

m

3

4n

5

d) Cn=2 (,2 - 1)2 = 4

prove the following: = 1 + 2 r + 3 r 2 + . . - + ( n + l ) r n + . . . , -1 < r < 1 a)

1

= 1 + 3 r + . . . + w +

..., -1 < r

b, (1 - r)3 5. Prove fie general binomial formula for negative exponents:

(1 - r)-k = 1

+ kr + rk(k12+. 2+1) .

4

.ss.

i

-

JGt,

<1

. . + k(k+ l ) . . . ( k + n 1.2...nR

+...

6. Assuming (as will be proved later) that sin x and cos x are representable for all x by the absolutely convergent series:

verify by means of series the identity: 2sinxcosx = sin2x.

410

Advanced Calculus, Fifth Edition

x

7. Let all sums be from 1 to co.If b, has been evaluated and {a,) is "close to (b,}", it may be of advantage to write C a n = C b, C(b, - a,). For example, to evaluate x ( 2 " I)-' = a,, one can choose C b, = 2-" and write

-

x

+

+

The last series converges much more rapidly than x ( 2 " I)-', since its test ratio has limit $, as opposed to $. One can repeat the process, using a, = (4" 2")-' and b, = 4-", and so on. a) Follow the steps described through one stage of the process, to find x ( 2 " I)-' = 0.7645. b) Evaluate C a, = x ( 2 " 9-l)-l, using C b, = C 2-" . C) Evaluate a, = C ( n 2 + 1)-l using b, = nP2, then using b, = finally using b, = n-6 (see Problem 1).

+

+

x

+

x

x

Remarks In general, for a, a rational function of n , as in (c), one can take b, to be the ratio of the leading terms in numerator and denominator. For a convergent series, the degree of the denominator must exceed that of the numerator by at least 2. By repeating the process several times one obtains

where c, is a rational function of n whose denominator has degree exceeding that of the numerator by at least m 1. The coefficients k2, k3, . . . can also be obtained as coefficients of the power series kixi of the rational function f (x) such that f (l/n) = a, for all n. Thus for a, = (n2 l)-l, f (x) = ( x - ~+ I)-' = x2 - x x6 - . . . as in Section 6.15 and 6.16. The sums n-p are values of the zeta function ( ( p ) (Section 6.16) and are tabulated (see for example the tables of Jahnke and Emde listed at the end of Chapter 7).

+ x +

+

6.11 SEQUENCES AND SERIES OF FUNCTIONS If for each positive integer n a function f,(x) is given, then the functions f,(x) are said to form a sequence of functions. It will usually be assumed that the functions are all defined over the same interval (perhaps infinite) on the x-axis. For each fixed x of the interval the sequence f,(x) may converge or diverge. The first question of interest for a particular sequence is the determination of the values of x for which it converges. Thus the sequence f,(x) = xn/2" converges for -2 < x 5 2 and diverges otherwise. Similar remarks hold for infinite series whose terms are functions: *- . : a

The nth partial sum of such a series is itself a function of x :

F;

Chapter 6 .Infinite Series

41 1

Convergence of the series is by definition equivalent to convergence of the sequence of partial sums; if Sn(x)converges to S(x) for certain values of x, then sTuDs301g

for this range of x. ,'r f r ;. L Examples of this have been considered already: the series l/nP, in which ' the terms are functions of p, and the geometric series C z , r n ,in which the terms are functions of r. The first converges for p > 1; the second for Ir I < 1. Other examples are the following:

EL,

which can be shown to converge for all x, for Ix I 5 1, and for all x, respectively. The tests for convergence of a series of functions are the same as those for a series of constants. For each particular choice of x (or whatever the independent variable may be) the series C un(x) is nothing but a series of constants. Thus when x ranges over an interval, the problem is really that of discussing the convergence of many (in fact, infinitely many) series. One could, of course, test a variety of x-values separately; this is rarely advisable, although a particular application might make it unavoidable. It will be seen that for many series it is possible to determine by a single analysis all the values of x for which the series converges. For power series, precise results follow. In many cases the ratio test alone gives" the answer; for example, for the series

$1

which represents e x , one has

Since this limit is 0 for each x , the series converges for all x.

Although it may thus be possible to prove that a series C un(x)converges for all x of a certain range, another question remains to be answered: Is the convergenceequally rapid for all x of the range? This question turns out to be important both for practical applications and for basic theory. The practical application is as follows. Suppose it is known that a series C un(x) converges for a < x < b to a function f (x). If the function f (x) is a complicated one, while the functions un(x) are simple, it would be a considerable simplification if we could replace the function f (x) by a partial sum Sn(x)= ul(x) . . un(x);for example, if the un(x)are of the form cnxn,we would be replacing f (x) by apolynomial. This amounts to "rounding off' the series

+ +

412

Advanced Calculus, Fifth Edition

4

.

of functions, just as we rounded off the decimal expansion of in Section 6.1. The procedure can be justified if the error made is below a prescribed value E that is, if If (x) - S,(x)l c 6 for a < x < b. However, it can happen, as will be seen below, that no matter how large n is chosen, the error If (x) - S,(x)l exceeds E for some x of the interval a < x < b. This does not contradict the fact that the series converges for each x; it is simply due to the fact that the number of terms needed to reduce the error below r varies from one x to another in such a manner that no one n will serve for the whole interval. When this difficulty arises, rounding off over the whole interval a < x < b cannot be justified. On the other hand, it may happen that, however E is chosen, an N can be found such that for each n 2 N, IS,(x) - f (x)l < E for all x of the interval a < x c b; if this is so, the series is termed uniformly convergent in the interval a < x < b. A uniformly convergent series can be rounded off to any accuracy desired:

x,"=,

DEFINITION The series un(x) is uniformly convergent to S(x) for a set E of values of x if for each r > 0 an integer N can be found such that

for n L N and all x in E. The essential idea is that N can be found as soon as E (or the number of decimal places desired) is known, without considering which x of the set E is going to be used. The set E must, of course, be a set of values of x for which the series converges, but it need not consist of all such values. In many cases it can be shown that only by restricting attention to a part of the range for which convergence is known can it be ensured that the convergenceis uniform. In the worst case, one could restrict attention to a finite number of values of x, for which the convergence must be uniform. The definition of uniform convergence is given in terms of series; it can equally well be given in terms of sequences: The sequence f,(x) is uniformly convergent to f (x) for a set E of values of x, if for each E > 0 an N can be found such that

for n 2 N and all x in E. Thus the uniform convergence of the series C un(x) is equivalent to the uniform convergence of its sequence of partial sums Sn(x). [Actually, sequences and series are simply two different ways of describing the same limit process; every series can be interpreted in terms of its partial sums, and every sequence S, can be interpreted as the partial sums of a series, namely, the - - . .] (S, series S1 (SZ- Sl) . . The most common case of uniform convergence is that in which the functions u,(x) are continuous in a closed interval a ( x 5 b and the series C u,(x) converges in this interval to a continuous function f (x). [It is shown below, in Theorem 3 1 of Section 6.14, that uniform convergence of the series guarantees the

+

+ +

+

Chapter 6 Infinite Series

Figure 6.8

Uniform convergence.

continuity o f f (x).] In such a case the absolute error JR,(x)Jin replacement off (x) by the partial sum Sn(x):

is continuous for a

<: x 5 b and has a definite maximum

in this interval:

-

R, is simply the worst absolute error occurring in the interval. Uniform convergence of the series for a 5 x 5 b is then equivalent to the statement: lim

n'bo

= 0;

that is, the series is uniformly convergent for a 5 x 5 b if and only if the maximum error approaches 0 as n + m (see Fig. 6.8). The new condition is simply a rewording of the preceding definition for the case considered, for convergence of R, to 0 is the same as the condition < E for n > N(E) and hence as the condition If (x) - S,,(x)l < E for n 2 N(r). A similar formulation can be given for the general case (when, for example, E need not be a closed interval). One simply defines as the least upper bound of If ( x ) - S,(X)( on the set E, that is, as the smallest number K such that If (x) Sn(x)l 5 K for all x in E; the existence of such a number is established in the same way as for the upper limit of a sequence (Section 6.3; see also Section 2.23). This reduces to the maximum of If (x) - S,(x)l on the set E if this function has a maximum; however, the function, even when continuous, need not have a maximum (for example, when E is an open interval; cf. Section 2.19). Just as for upper limits, the least upper bound may be +m. Uniform convergence of C un(x) to f (x) in the set E is then equivalent to the statement: En -, 0 as n + oo,that is, to the statement:

xn

x,

lim 1.u.b. I f (x) - S,(x)l = 0,

n-tm x In E

where "1.u.b." stands for "least upper bound."

(6.27')

413

414

Advanced Calculus, Fifth Edition

Figure 6.9

Sequence of partial sums of the geometric series.

EXAMPLE 1 The geometric series

*

x n . The series converges for

-1 < x < 1.

The nth partial sum is S,(x) = 1 + x

+ .. . + xn-'

1-xn =1-x'

and the sum is

these are plotted in Fig. 6.9. The remainder R,(x) = S ( x ) - Sn(x) satisfies the equation:

We consider first a closed interval: - 5 x 5 in this interval. We find

i. The convergence is uniform

Chapter 6 Infinite Series for when x = $, the numerator Ixn1 has its largest value'arid the denominator 11-2kl, has its smallest value in the interval. Accordingly,

-

--

lim R, = lim

.

n+w

n-rw

1

= 0; 2"-1

the maximum absolute error tends to 0 as n -+ oo,and the convergence is uniform. A similar reasoning applies to eachclosed interval -a 5 x 5 a (0 < a < 1 ) within the-interval of convergence. The worst absolute error occurs for x = a , and its value is R, = a n / ( l- a ) ;this tends to 0 as n + oo,so that the convergence is uniform in the interval. One might expect from this that the convergence would be uniform over the entire open interval - 1 < x < 1 . This is not the case. Indeed, for each n the absolute error I Rn( x ) (is unbounded for - 1 c x < 1 , since

xn

is always +oo! We can see the difficulty in detail by The least upper bound asking how many terms are needed to compute the sum with an absolute error less than E = 0.01, for example. For x = 0, one term is sufficient; for x = 0.5, one must have (0.5)" 0.5

-< 0.01

or 2"-' > 100;

hence n must be at least 8. For x = 0.9, one must have

this requires n > 65, as a computation with logarithms shows. As x increases toward 1 , more and more terms are needed; one can easily show that, as x += 1 , the number of terms required approaches oo (see Problem 2 following Section 6.9). It appears from this discussion that the difficulty is due to the fact that the sum S(x) = 1 /(1 - x ) becomes infinite as x approaches 1. However,the convergence is not uniform in the interval - 1 < x 5 0. For the absolute error I Rn(x)(lies between 0 and f in this interval; this is suggested graphically in Fig. 6.9 and can be proved strictly by calculus. Since I Rn(x)l+ as x -t - 1 , the least upper bound is always this is not the maximum absolute error, since x = - 1 is excluded from the interval under consideration. Accordingly, i?,,cannot converge to 0 as n + oo,and the convergence is nonuniform for - 1 < x 5 0. Again we can verify that the number of terms needed to compute S(x) with an error less than E = 0.01 approaches oo as x + -1.

xn

i;

EXAMPLE 2 The sequence fn(x) = xn converges to 0 for - 1 < x < 1 and to 1 for x = 1 ; it diverges for all other values of x. The convergence for - 1 < x < 1 is

415

416

Advanced Calculus, Fifth Edition not uniform. For 0 Ix < 1 the error is precisely x n ; for this to be less than E , one must have log r X " < E or n > - . log x As x approaches 1, the value of n required approaches +m. For x = 1 the error is always 0. Successive members of the sequence f,(x) are plotted in Fig. 6.10 (in Section 6.14). Since the worst errors occur near x = f1, one can obtain uniform convergence by restricting x as in the preceding example: -0.5 5 x 5 0.5.

6.13 WEIERSTRASS M-TESTFOR UNIFORM CONVERGENCE The following test is adequate for determining the uniform convergence of a large number of familiar series. THEOREM 30 (Weierstrass M-Test) Let C,"=, u,(x) be a series of functions all defined for a set E of values of x . If there is a convergent series of constants C z ,M, , such that l u , ( x ) l ~ M , forallxinE,

then the series Czl U , ( X ) converges absolutely for each x in E and is uniformly convergent in E. Proof. The Weierstrass M-test is first of all a comparison test. For each fixed x , each term of the series C lu,(x)l is less than or equal to the nth term M, of the convergent series C M , . Hence by the comparison test (Section 6.6, Theorem 12) the series C u,(x) is absolutely convergent. But the comparison series C M, is the same for all x of the range considered. It is from this fact that the uniform convergence follows. For if R, = S - S,(x) is the remainder after n terms of the series C u,(x), then

as in Theorem 22 of Section 6.9. In other words, if T, denotes the remainder after n terms of the convergent series C M,: T, = Mn+l

+ Mn+2+ . ' - ,

then IRn(x)I ITn. Since M, is a series of constants, for each given E > 0,N can be found such that T, < E for n 1 N; for this same N, one has

1

IR,(x)l IT, < E forn 2 N. Since N does not depend on x , but only on r , the uniform convergence has been established. At the same time it has been shown that T, serves as an upper estimate for the error committed in using only n terms of the series C u,(x), regardless of the x . The error will, of course, vary from one x to another, but the upper estimate is the same for all x .

t

EXAMPLE 1

Ezl5.Here the ratio test gives ' ' "

pimlzl an+l

l---l

xn+l

= nlim + w ( n + 1)2

n2 x n = nlim -w

417

Chapter 6 Infinite Series

.

''*

*,

3fl1 ,

.- -

,?a .<

II( (a = 1x1. ? 0 1x1 . (n+ n2

,

,

Hence the series converges for (x( < 1 and diverges for 1x1 1 . For x = f1 the series converges by comparison with the harmonic series of order 2:

Hence the series converges for - 1 5 x I:1 . The convergence is uniform for this range since the comparison

holds for all x of the range and the series C M,, converges.

EXAMPLE 2

Eml y.This series converges uniformly for all x,since'"

.'

for all x , and the series C ( 1 / 2 " ) is convergent.

1 B1.b EXAMPLE 3 $. The ratio test shows, as in Example 1 , that the series converges for - 1 < x < 1 and divergesfor Ix I > 1 . For x- = 1 the series is the divergent harmonic series; for x = - 1 the series is a convergent alternating series. One might try to prove uniform convergence for 0 5 x < 1 by using the inequality

xm,

'.CIAdl.

'e,

ELl

and the fact that the series x n converges. However, this feasoning is irlmect? since the comparision series C x n depends on x and is not a series of constants as required in Theorem 30. As it happens, the given series is not uniformly convergent for0 ~x < 1 . A

4:

PROBLEMS 1. Determine the values of x for which each of the following series converges:

418

Advanced Calculus, Fifth Edition 2. Prove that each of the following series is uniformly convergent over the set of values of x given: (tanh x)" a) C2=l n. -1 5 x r 1 b)CZ ! allx

5,

C)

C ~;;r+ I ll d

L) c = ~ $ , g)

+

sin nx

x~log$ 1 1 f) C E l n x n , -2 I X L 2 h) C E l n x n , -a 5 x 5 a , a c 1 d)

x

-1rxrl.

CE,n x n , -0.9

5 x 1 0.9

xzl

#I

3. Prove: If u n ( x )is uniformly convergent for a I: x 5 b, then the series is uniformly convergent in each smaller interval contained in the interval a 5 x 5 b. More generally, if a series is uniformly convergent for a given set E of values of x , then it is uniformly convergent for any set El that is part of E. 4. Prove: If C z l vn(x)is uniformly convergentfor a set E of values of x and I u ~ ( x ) I 5 vn(x) for x in E, then CF=, un(x)is uniformly convergent for x in E. 5. Prove: If 0 < u n ( x ) < l l n and u , + ~ ( x )i u n ( x ) for a 5 x 5 b, then the series (- 1)"un(x)is uniformly convergent for a 5 x 5 b. 6. Prove: If the series C z l M,, of constants Mn is convergent and I fn+,(x) - fn(x)l 5 M,, for x in E, then the sequence f n ( x ) is uniformly convergent for x in E. 7. Prove that the following sequences are uniformly convergent for the range of x given (cf. Problem 6): a) %, 0 5 x 5 1 b) $, - 1 5 x 5 1

xz,

C)

d) np, 21 ~ x 1 1

' ~ g ( ' ~ + " ~ 1) ,5 x 5 2

(B.

6.14 PROPERTIES OF UNIFORMLY CONVERGENT SERIES AND SEQUENCES Let C u n ( x ) be a series of functions, each of which is defined for a 5 x 5 b. Let it be assumed further that this series converges to a sum f ( x ) for a 5 x 5 b, so that one has

One can then ask questions such as the following: If each function u n ( x ) is continuous, is the sum f ( x ) continuous? If each u n ( x )has a derivative, does f ( x ) have a derivative? The following theorems answer such questions.

THEOREM 31 The sum of a uniformly convergent series of continuous functions is continuous; that is, if each u n ( x )is continuous for a p x 5 b, then so is f ( x ) = C z lu,(x), provided that the series converges uniformly for a 5 x 5 b. Proof. that

Let xo be given, a 5 xo 5 b, and let 6 > 0 be given. We then seek a S such

1f

( x ) - f (xo)l < E

when Ix

- xol

<6

and x is in the given interval. We choose N so large that

I

-f

<

1

5 , a 5 x 4 b,

n

t N.

(6.28)

.

41 9

Chapter 6 Infinite Series

Figure 6.10

The sequence f , , ( x ) = x".

+ +

where Sn(x) = u l ( x ) - - . u,(x); this is possible since the series is uniformly convergent. The function SN(x),as sum of ajnite number of continuous functions, is itself continuous. One can hence choose a 6 such that IS&)

1

- SN(XO)Ic j . ~ for I X

- XOI < 8 .

+

+

(6.29)

By (6.28), one has

Hence

If

( x ) - f (xo)l = If ( x ) - S N ( X ) SN(X)- SN(XO) SN(XO) -f(xo)~

1 1 1 + -E -6 = E , for J X - XOI < 6 , 3 3 3 by (6.29) and (6.30). Thus continuity is proved. <

-E

1

+

'

\

Remark 1. The property of convergence alone, for a series of continuous functions, does not guarantee continuity of the sum. This is seen by the example:

Here the nth partial sums form the sequence Sn(x)= xn plotted in Fig. 6.10. As was pointed out in Section 6.12, this sequence does not converge uniformly. The sum of the series is 0 for 0 5 x < 1 and 1 for x = 1; there is a jump discontinuity at x = 1. Remark 2. Theorem 3 1 can be interpreted in terms of sequences as follows: If Sn(x) is a sequence of functions all continuous for a (x 5 b and this sequence converges uniformly to f ( x ) for a 5 x 5 b, then f ( x ) is continuous for a Ix 5 b; furthermore, if a 5 xo 5 b, then

lirn s.(x)] = lirn n+m

[ lim $ ( x ) ] . x+xo

1

(6.31)

420

Advanced Calculus, Fifth Edition The left-hand side is precisely limx,,o f ( x ) , and the right-hand side is limn,, Sn(xo)= f (xo),by the continuity of S,(x); Eq. (6.31) is simply the assertion that

that is, that f ( x ) is continuous at xo. Accordingly, as Eq. (6.31) shows, uniform convergence permits one to interchange two limit processes. THEOREM 32 A uniformly convergent series of continuous functions can be integrated term by term; that is, if each u n ( x )is continuous for a 5 x ( b, then

Ib

b

b

f ( x ) d x = l ul(.r)dx+L ut(x)dx+...+

Ib

u,(x)dx+...

.

(6.32)

1.

Proof. As in the preceding proof, let Sn(x)be the nth partial sum of the series C un(x).Then

b

S n ( x ) d x converges to To prove (6.32), one must show that the sequence lob f ( x )d x , that is, for each z 0 an N can be found such that

To establish this, we choose N so large that €

If ( x ) - &(x)I < b-o ,

n > N , a ~ x ~ b ;

this is possible because of the uniform convergence. Hence

Ilb

f(x)dx -

lb

I

€

Sn(x)dxl = I l b [ f ( x )- & ( x ) ] ~ x5 -. b-a

(b-a)=€,

.d

n L N, by the rule (4.6) in Section 4.1. Theorem 32 is thus established.

•

Remark 3. This theorem can be formulated in terms of sequences and interchange of limit processes as in the preceding Remark 2. EXAMPLE In Section 6.12 the series CEO xn was shown to converge uniformly to 1/(1 - x ) in each interval -a 5 x 5 a , where a < 1. On integrating the equation

from 0 to xl, one finds

Chapter 6 Infinite Series

421

This holds for every xl between -1 and 1 (the ends excluded), so that one can write 1 1 =x+-x2+...+-+..., -1 < x < 1. (6.33) l-x 2 n+l The same procedure could be phrased in terms of indejnite integrals. Thus log-

1 1 1 3 =x+-x2+-x c. 1-x 2 3 n+l * Of course, c is no longer arbitrary; by choosing x = 0 we recognize that c = 0,so that (6.33) is again obtained. log-

+ . . a + - + . . . +

.t

THEOREM 33 A convergent series can be differentiated term by term, provided that the functions of the series have continuous derivatives and that the series of derivatives is uniformly convergent; that is, if uL(x) = dun/dx is continuous for a 5 x 5 b, if the series C,"=, un(x)converges for a 5 x 5 b to f (x), and if the series CFl uL(x) converges uniformly for a 5 x 5 b, then m

The derivatives at a and b are understood as right- and left-sided derivatives, respectively. Proof of Theorem 33. Let g(x) be the sum of the series of derivatives:

Then g(x) is continuous by Theorem 31, and by Theorem 32,

Hence

1

= x u n ( ~ ~C) ~- n ( a ) . n=l n=l

t r

-9

Accordingly,

If both sides are differentiatedwith respect to XI,the fundamental theorem of cdculus (Section 4.1) gives

422

Advanced Calculus, Fifth Edition In other words,

% -

xz,

t

.

7"-

xzl

un(x) and vn(x) are uniformly convergent for THEOREM 34 If a 5 x 5 b and h ( x ) is continuous for a 5 x 5 b, then the series

are uniformly convergent for a 5 x 5 b. Proof. Let f ( x ) and g ( x ) be the sums of C un(x)and C vn(x),respectively; let Sn(x)and Q n ( x )be the corresponding partial sums. The nth partial sum of C ( u n+v,) is Sn Qn. Let N be chosen, for given 6 , so that

+

ISn(x) - f (x)l <

1 26'

IQn(x) - g(x)l <

1 I€,

forn 2 N anda 5 x 5 b.Then

+

+

Thus C ( u n u,) converges uniformly to f (x) g(x). A similar proof applies to the difference. Since h ( x ) is continuous for a 5 x 5 b, it is necessarily bounded: Ih(x)l 5 M for a 5 x 5 b. Hence Ih(x)Sn(x)- h ( x ) f (x)l = Ih(x)llSn(x)- f (x)l < M

I'

. 26

?

< Me

for n 2 N as before. This shows that the series C h(x)un converges uniformly to h ( x )f ( x ) . It should be noted that actually only the boundedness of h ( x ) was required.

By a power series in powers of x is meant a series of the form

where co, c l , . . . , c,, meant a series:

.. . are constants. By a power series in powers of ( x - a ) is

00

E c , ( x - a)n = co + c l ( x - a ) + n=O

... +cn(x -a)"

+ . . a .

(6.36)

. ,. Divergence

Rgure 6.11

I a - r*

Convergence a

Chapter 6 Infinite Series I

Divergence X ,

a + r*

Interval of convergence of a power series.

By a power series in negative powers of x is meant a series:

The term power series alone usually refers to (6.36), of which (6.35) is a special case (a = 0). The substitutions t = x - a or t = l/x can be used to reduce series of form (6.36) or (6.37) to the form (6.35). The power series (6.36) converges when x = a . It can happen that this is the only value of x for which the series converges. If there are other values of x for which the series converges, then it will be seen that these form an interval, the "convergence interval," having midpoint x = a . This is pictured in Fig. 6.1 1. The interval can be infinite. These properties are summarized in Theorem 35.

THEOREM 35 Every power series co

+ cl(x - a ) + ... + cn(x -a)" + ..-

has a "radius of convergence" r* such that the series converges absolutely when Ix - a1 < r* and diverges when (x - a ( > r*. The number r* can be 0 (in which case the series converges only for x = a), a positive number, or oo (in which case the series converges for all x). If r* is not zero and rl is such that 0 < rl < r*,then the series converges uniformly for Ix - a1 5 r l . The number r* can be evaluated as follows:

r* = i m n-'co

r* = lim n+m

1% 1,

if the limit exists,

1 -

gq'

if the limit exists,

r* = lim

1

Cn+l

and in any case by the formula:

n-m

m'

Proof. We begin by considering the case in which the limit (6.38) exists, as this at once suggests the general situation. The ratio test, applied to the given series, proceeds as follows:

-

lim

C,+I(X- a)"+] C,(X

- a)"

423

424

Advanced Calculus, Fifth Edition The series converges absolutely when

I X - al r*

e l ,

thatis,

Ix-al
1.

.m,.1

and diverges when that is,

I X - a1 > r*.

If r* = 0, the test shows divergence except for x = a . If r* = oo, so that

the test shows that the series converges absolutely for all x . The case in which the limit (6.39) exists is treated in the same way by the root test. This case is included in the final formula (6.40), which follows at once from Theorem 21(a) of Section 6.8, for

The series converges absolutely when the upper limit is less than 1, that is, when 1 = r*, I X -a1 e lim n+m

and diverges when Ix - a 1 > r*. Thus however it may be determined, there is a number r*,0 5 r* 5 oo, with the properties described. It remains to prove that if 0 < rl < r*,then the series converges uniformly for Ix - a1 5 r l . This follows from the M-test, with Mn = Icnlr;, for

and this converges, since Ixl - a 1 = rl < r*. For Ix

- a 1 5 rl,

so that the convergence is uniform. The main idea of this proof is emphasized in Section 6.12-the slowest convergence of a power series occurs toward the ends of the interval of convergence. It should be emphasized that when r* is a finite positive number, the series may converge or diverge for each of the end-values x = a r*,x = a - r*.These must be investigated separately for each series.

+

EXAMPLE 1

x E l 5.Here cn = 1/ n 2 , and (6.38) gives

+

(n r* = lim -= 1: similarly, (6.39) gives

"

%

Chapter 6 Infinite Series

,

425

by Theorem 4 of Section 6.4. Hence the series converges absolutely for -1 < x < 1 and diverges for Ix 1 > 1. For x = f 1 the series converges absolutely, since

IT\

1

(fl)"

0

=

r

2.

:

f

:&iw rn In fact, the series converges absolutely and uniformly for - 1 5 x 5 1, since

in this range.

.

P f

.

xzl

$.The ratio formula and root formula apply as in Example 1 EXAMPLE 2 to give r* = 1. Hence the series converges absolutely for 1x1 < 1 and diverges for 1x1 > 1. For x = 1 the series is the harmonic series and diverges; for x = -1 the series converges conditionally by the alternating series test. Hence the series converges for - 1 5 x < 1 and diverges otherwise. It should be noted that Example 2 is obtained from ~ x a m p l e1 essentially by term-by-term differentiation:

an extra factor of x is needed. The effect of term-by-term differentiation is thus to" multiply each term by n (while lowering the degree of xn by 1). This cannot affect .* the radius of convergence, since

However, it does slow down convergence at the ends of the interval, as the second example shows. If one differentiates the series of Example 2, one obtains the geometric series:

which converges for - 1 < x < 1 and diverges at the ends of the interval.

THEOREM 36 A power series represents a continuous function within the interval of convergence; that is, if r* is the radius, then

is continuous for a - r* < x < a

+ r*. +

Proof. By Theorem 35 the series converges uniformly for a - rl 5 x (a rl, rl < r*. Each term cn(x - a)" of the series is continuous for all x ; hence by Theorem 31, f ( x ) is continuous for a - rl ( x ( a r l . This holds for every rl e:. between 0 and r*. Accordingly, f (x) is continuous for a - r* < x < a r*.

+

+

.-

426

Advanced Calculus, Fifth Edition Remark. If the series converges at either end of the interval, then f ( x )is continuous at the end concerned. For a proof refer to Chapter V of the book by Knopp listed at the end of the chapter. 0

THEOREM 37 A power series can be integrated term by term within the interval of convergence; that is, if 00

kt

f ( ~ ) = C c ~ ( x - a ) a~ -,r * < x < a + r * , n=O

.-

L,

then for a - r*
hii;

or in terms of indefinite integrals, U~

" n l l

TL,

u-r

Proof. By Theorem 35 the series converges uniformly in each interval a - r l (x 5 a r l . It is therefore uniformly convergent for each interval x l Ix 5 x2, for the latter interval will be included in the former for appropriate choice of rl. One can now apply Theorem 32 to justify term-by-term integration. The indefinite integral can be regarded as a special case of this, for if xl = a , then

+

3

13,

and

by the fundamental theorem of calculus; that is,

The number x2 can be chosen anywhere in the interval a - r* < x < a accordingly, the last equation is valid for all x of this interval. x .

+ r*, and •

Chapter 6 Infinite Series THEOREM 38 A power series can be differentiated term by term within the interval of convergende; that is, if J

dtrw

. ~ ~ q s ~

>tt=hf

00

f(x)=Ccn(x-a)",

a-r* < x < a + r * ,

then 00

f l ( x ) = x n c n ( x - a)"-',

a - r* < x < a

+ r*.

Proof. The general coefficient of the differentiated series is nc, [or, more precisely, (n l ) ~ , + ~The ] . extra factor of n has no effect on the upper limit of the nth root, as was pointed out in connection with Example 2. Hence the differentiated series has the same radius of convergence r*; this series is then uniformly convergent in each interval a - r1 5 x 5 a rl ,rl < r *. Hence by Theorem 33,

+

+

+

for each x of such an interval. Every x such that a - r* < x < a r* lies in such an interval, so that the result is justified for all x within the interval of convergence. Application.

From the relation

one now obtains by successive differentiation the relations

The general case can be written thus:

- l < x < l , k = l , 2 , 3 ,.... This is termed the binomial theoremfor negative integral exponents. Equation (6.41) is actually valid for every real number k, as will be seen.

427

428

Advanced Calculus, Fifth Edition

6.16 TAYLOR AND MACLAURIN SERIES Let f ( x )be the sum of a power series with convergence interval a - r* < x < a + r* (r* > 0): co

f ( x ) = ~ c n ( x - a ) " , a-r* 4 x
(6.42)

This series is called the Taylor series off ( x )at x = a if the coefficients c, are given by the rule: cg

f '(a)

= f ( a ) , cl = -, c2=-

l!

f " ( a ) ... , 2! '

Cn

f '"'(a) " ' =n! 9

9

so that

THEOREM 39 Every power series with nonzero convergence radius is the Taylor series of its sum. Proof. Let f ( x )be given by (6.42). Then by repeated differentiation, on the basis of Theorem 38, one finds +cn(x -a)" f ( x ) = co +c1(x - a )

+ f ' ( x )= cl + 2c2(x - a ) + . . . + n cn(x - a)"-' + . . . , f " ( x )= 2 ~ 2+ 6c3(x - a ) + . . + n(n - 1) cn(x +. -, + . . a ,

.. ..

+ 1

.

Here all series converge for a - r* < x < a r*. If one now sets x = a , one finds f ( a ) = co, f ' ( a )= cl, f " ( a )= 2c2, . - . , f ("'(a)= n!cn,. ... This gives co = f ( a )and I

as asserted. In the case in which a = 0 the expression (6.43) for the Taylor series of f ( x ) becomes f'(o)x + f"(o)x2+...+ f '"'(0)~" (6.44) f ( x )= f (0) l! 2! n! This is called the Maclaurin series of f ( x ) .For many purposes it is easier to employ. The substitution t = x -a reduces the general Taylor series to the form of Maclaurin series.

+

THEOREM 40 If two power series

+

a

*

-

.

nL:

429

Chapter 6 Infinite Series

have nonzero convergence radii and have equal sums wherever both series converge, then the series are identical; that is, cn=Cn,

n=0,1,2

,....

f

Proof. Let the convergence radii be r* and R*, respectively, where 0 < r* 5 R*. Then, by assumption,

By the preceding theorem, one must then have co = Co= f ( a ) , and

Thus the coefficients are equal for every value of n.

COROLLARY If a power series has a nonzero convergence radius and has a sum that is identically zero, then every coefficient of the series is zero. I

PROBLEMS 1. a) Obtain the Maclaurin series

1 1 xn log-=x+-x2+...+-+... , -1 < x < 1 , 1-x 2 n by integration of the series for 1/(1 - x ) . Verify that (6.43) holds. b) Show that the series converges for x = - 1 and hence prove (on the basis of the remark preceding Theorem 37) that

2. Prove (6.41) by induction. 3. a) Expand l / x in a Taylor series about x = 1. [Hint: Write 1/[1 - ( 1 - x ) ] = 1 / ( 1 - r ) and use the geometric series.]

b) Expand l / ( x +2) in a Maclaurin series. [Hint: Write l / ( x + 2) = 1/(2[1- (- !jx)]]= ' $ ( l / ( l - r)).l c) Expand 1 /(3x 5 ) in a Maclaurin series. d) Expand 1/ ( 3 x 5 ) in a Taylor series about x = 1. [Hint: Write 1 /(3x 5 ) = 1 / [3(x - 1 ) 81 = ; / [ I $ ( x - 1)I.l e) Expand l / ( a x b ) in a Taylor series about x = c, where ac + b # 0,a # 0. f ) Expand 1 / ( 1 - x 2 ) in a Maclaurin series. g) Expand l / [ ( x- 2)(x - 3)] in a Maclaurin series. [Hint: Write l / [ ( x - 2)(x - 3)] = A / ( x - 2) B / ( x - 31.1 h) Expand 1/x2 in a Taylor series about x = 1. [Hint: Write 1/x2 = 1 / [ 1 - ( 1 - x)12 = 1/(1 - r)' and use (6.41).] i) Expand 1/(3x f 5)2 in a Taylor series about x = 1.

+

+

+ + +

+

+

L

430

AdvancedCalculus, Fifth Edition

+

+

j) Expand l / ( a x blk in a Taylor series about x = c, where ac b # 0, a # 0, and k = l , 2 , .... 4. Let f ( x ) = C z l $. a) Show that f ( x ) is defined for all x . b) Evaluate (approximately where necessary) f (O), f (1). f '(0),f '(1),f "(0). c ) Obtain Maclaurin series for f '(x),f " ( x ) . 5. Let y = f (x) be a function (if there is one) such that f ( x ) is defined for all x , f ( x ) has a Maclaurin series valid for all x , f (0)= 1 and dyldx = y for all x . Show that necessarily f(x)= 1+x+x2/2!+-..+xn/n!+... .A and that this function satisfies all requirements. [It will be shown later that f (x'j = ex.]

6.17 TAYLOR'S FORMULA WITH REMAINDER The preceding discussion concentrated on the power series, rather than on the functions that they represent. The opposite point of view is also of fundamental importance, the first question being the following: Given a function f ( x ) ,a < x < b, can f ( x )be represented by a power series in this interval? If f ( x )can be so represented, f ( x )is termed analytic in the given interval. More generally, f ( x )is termed analytic for a < x < b if for each xo of this interval, f ( x )can be represented by a power series in some interval xo - 6 < x < xo 6 . Most of the familiar functions-polynomials, rational functions, ex,sinx, cos x, log x , A-and those constructed from them by operations of algebra and by substitutions are analytic in every interval in which the function considered is continuous. The exceptions are not too difficult to recognize. Thus 0= Ix 1 is continuous for all x but has a discontinuous derivative for x = 0. Hence the function cannot be analytic in an interval containing this value. The function f ( x ) = e-'fX2is defined and continuous for all x other than 0. If one defines f (0)to be 0, the function becomes continuous for all x and can in fact be shown to have derivatives of all orders for all x (Problem 6 following Section 6.18).However, the function fails to be analytic in any interval containing x = 0.For example, one finds that for this function, f (0) = 0, f '(0)= 0 , . , . , f(")(O) = 0 , . . . , so that the Maclaurin series would reduce identically to zero. The series converges but does not represent the function. A satisfactory theory of analytic functions is most easily achieved with the aid of complex variables. The reader is referred to Chapter 8 for this topic. However, the following theorem does prove useful in establishing analyticity without recourse to complex numbers.

+

)1

THEOREM 41 (Taylor9s ]Formula with Remainder) Let f ( x ) be defined and continuous and have continuous derivatives up to the (n 1)st order for a - ro < x < a ro. Then for each x of this interval except x = a ,

+

+

f '(a) - a ) . - . + f '"'(a)( X - a)" f(x) = f ( a ) + -(x 1 n!

+ f(n("+"(xl) + l)!

(X

- a)"" . .. • %

for some xl such that a < xl < x or (if x < a ) x < xl < a .

It should be noted that for n = 0 the theorem reduces to the Mean Value theorem: f ( x ) = f ( a ) f '(xl)(x- a ) . For general n it gives an expansion identical with

+

431

Chapter 6 Infinite Series

-

the Taylor series up to the tin (x n r , with the rest of the series replaced by a single term. We prove the theorem for n = 1, leaving the general case as a problem (Problem 3 following Section 6.18). Let x2 be a fixed number in the given interval, x2 # a . and let

Then F is defined and continuous forx in the given interval, and F(a) = 0, F(x2)= 0. Hence by the Mean Value theorem, F1(xl)= 0 for some xl between a and x2. But a calculation shows that

The equation F1(xl)= 0 thus becomes the equation

--

I

,

r!n,tc

hr.

If x2 is now replaced by a variable x, one has the desired result: f(x) = f (a) +(x -a)fr(a)

+ 1 fr'(x~)(x-ad.

Now let f (x) be a function having derivatives of all orders in the given interval so that one can form all terms of the (hypothetical) Taylor series of f about x = a . Although this series may fail to converge, except for x = a , and, even if convergent, may fail to have f (x) as sum, one can nevertheless write for each n:

where R, is the remainder term:

Since xl is not explicitly given, the remainder is not explicitly known. However,

'

Eq. (6.45) can often be used to obtain an upper estimate for I R, I. From this estimate, one may then be able to show that lim R, = 0

Y.T'%WAY

n+m

for all x of the chosen interval. If this has been demonstrated, then one concludes that

that is, that f (x) is represented by a Taylor series over the given interval and is analytic; at the same time, one has proved convergence of the series.

,

432

Advanced Calculus, Fifth Edition EXAMPLE

Let f ( x ) = e x . Then, for a = 0 and x > 0,

Hence O
exn+'

+

(n l ) !' This implies that R, is less than the nth term of the series

which converges, by the ratio test, for all x . Thus one has '

*-3

lim -- 0 ,

n+m

(n

+ I)!

and accordingly, lim R, = 0. A similar discussion applies for x < 0.Accordingly, ex can be represented by a Taylor series: X

x2

l!

2!

e'= 1 + - + - +

xn xn ...+,+...=En?, allx. n. OC

n=O

In a similar manner, one can prove that the following expansions are valid:

-X - - +x3- + .x5 ..+

slnx=

l!

3!

x2 2!

5!

x4 4!

cosx=l--+-+...

(- 1)"+lx2"-'

(2n - I)!

+.

1>"xk + (-(2n)! + ...,

,

.,

all x;

all x ;

- 1 < x < 1, m any real number.

.

,

[For the derivation of (6.49), see CLA, Section 6-20.] A variety of other expansionscan be obtained from these by appropriate substitutions and combinations. The series in (6.49)becomes the geometric series when m = - 1 and x is replaced by - x . This series can be used, as indicated in Problem 3 following Section 6.16, to obtain expansions of other rational functions.Then by differentiation and integration, further results can be obtained. EXAMPLE 1 With m = - 1 and x replaced by x 2 , (6.49) becomes 1

--

1 - x 2 + . . .+(-1)"x2"+

1 +x2 If this is integrated, one finds

...,

-1<X<1.

(6.50)

I

.

+ P ) ,one finds

EXAMPLE 2

Since coshx = ;(ex

EXAMPLE 3

Since sin x cos x = sin 2x, one finds

t:

5

sinxcosx = - - - -+...+ 2 1 ! 23x3 3!

433

Chapter 6 Infinite Series

I

(- 1)"-122"-l*Zn-1

(2n - l)!

+ . .} .

(6.53)

Remark. Since the remainder formula does provide a method for estimating R,, it enables one to estimate the error committed in computing the sum of a power series. Thus if f (x) is known to be analytic in a given interval and it is known that in this interval,

'

:.

for a certain constant Mn+1,then

This formula can be added to those developed in Section 6.9. As Theorem 41 shows, one does not require analyticity of the function f (x) in order to be able to apply the remainder formula; f (x) need only have continuous derivatives through the (n 1)st order. Actually, the (n 1)st derivative need only exist between a and x, continuity not being required. Thus in principle, one can use ' the formula as a method for evaluating a nonanalytic f ( x ) by a Jinite series, with remainder estimated by (6.54).

+

+

Four other operations by which new Taylor series can be obtained are described by the following theorems. i

THEOREM 42

Convergent power series can be multiplied; that is, if -T

m

M

'

434

Advanced Calculus, Fifth Edition are power series with convergence radii r,* and r,*,respectively, 0 < r,* 5 rf, then

where

kn = coCn

+ clCn-1 + ~2Cn-2+ + cn-1C1+ cnCo. '''

This is simply an application of the Cauchy product rule (Section 6.10) to the absolutely convergent series for f ( x )and F ( x ) . THEOREM 43 Convergent power series can be divided, provided that there is no division by zero; that is, if f ( x ) and F ( x ) are given as in Theorem 42 and F ( a ) = Co # 0,then -

for some positive number r;, where the pn satisfy the equations Cn

= poCn

+ plcn-1 + . - .+ pn-lC1 + pnCo.

.

.

3

(6.55)

The rule (6.55) expresses the fact that the series pn(x - a)" multiplied by the series C Cn(x - a)" gives the series C cn(x - a)".A proof of this theorem and a more precise determination of r; require complex variables; this is taken up in Chapter 8. It should be noted that the Eq. (6.55) are implicit equations for the coefficients pn : co = poco,

Cl

= poC1

+ p,Co, . . .

These can be solved in turn to obtain as many coefficients as are desired:

it will usually be difficult to obtain a formula for the general coefficient pn. THEOREM 44 A Taylor series with constant term a can be substituted for the variable x in a Taylor series about x = a ; that is, if

have nonzero convergence radii r,* and r,*, respectively, and Ig(x) - a ( < r,* for Ix - bl < r ~where , r2 5 rl*, then

where the coefficients qn are obtained by collecting terms of same degree.

I:I

Chapter 6 Infinite Series

435

An example will make clear the rule of formation of coefficients:

>no In this case, r2 = ca. The proof of the theorem is simplest in terms of complex variables and will not. be considered here. The theorem is a special case of a theorem of Weierstrass on double series, for which the reader is referred to the book by Knopp given at the end of the chapter.

THEOREM 45 A power series can be inverted, provided that the first-degree term is not zero; that is, if

and cl # 0, then there is an inverse function

I r;

The coefficients b, are determined from the identity

Again, an example will clarify the method of determining coefficients. From the series (6.5 1):

arc tanx = x - -1x 3 -I--.., 3 one can seek to determine a series for x = tan y :

I

I

I

436

Advanced Calculus, Fifth Edition Hence

so that

For a proof of the theorem, one is referred to page 184 of the book of Knopp referred to earlier. This last theorem and the one on division suggest a principle that has a broad field of applications: In order to determine a function that is to satisfy a given condition, one can postulate that the function is expressible by a power series and then try to determine the coefficients of such a series in order to satisfy the given condition. If such a series can be found, one can then investigate the convergence of the series and determine whether it actually defines a function satisfying the given condition. Remark. The operations on power series can be reduced to calculations with n x n matrices, with n chosen sufficiently large. For details one is referred to Chapter 1 of the book by Henrici given at the end of the chapter.

PROBLEMS 1. Obtain the following Taylor series expansions: a) sinh x =

b) cos2 x = 1

bO

-,

X2n-I

+Czl

(_l)n22n-lX2n

(2n)!

(-l)n+122n-IX2n

Cz, -l)n+l(X d) log x = Czl n c) sin2x =

all x

e ) G = l - $ - = x 1

, all x , all x

- l)n 2

, Ix-lI
2

f ) -m 1 _ -1 + + + V x 24 + 21 L p x 62.+3!. . . , 1.3.5~'

g) a r c s i n x = x + & + & $ + m r + . . . ,

1x1 < 1 1x1 < 1 1x1 < 1.

2. Find the first three nonzero terms of the following Taylor series: b) tan x about x = 0 a) ex sin x about x = 0 d) log (1 - x2) about x = 0 C) log2 (1 x) about x = 0 f ) etanxabout x = 0 e) x 3 + 3 x + 1 a b o u t x = 2 h) y = tanhx about x = 0 g) y = sinh-' x about x = 0 j) y = logsecx aboutx = 0 i) y = tanh-' x about x = 0 3. Prove Taylor's remainder formula (Theorem 41) for general n. [Hint: Replace the function F ( x ) by the function G(x) - [(x2 - x)/(xz - a)lnG(a),where

+

4. Show that the remainder term Rn of Taylor's formula can be written in the integral form:

.

Chapter 6 Infinite Series

437

[Hint: Use induction. For n = 0 the formula becomes the known rule

+

Suppose the formula is true with n replaced by m and prove it is true for m 1; for this step, integrate the remainder term by parts taking u = f(m+')(t),dv = (x - t)m dtlm!.] 5. Evaluate to threc decimal places: 0.5

a) d j ePxzdx

"J,

, , dx

6. Let f (x) = e-'1"' for x # 0 and let f (0) = 0. a) Prove that f (x) is continuous for all x. b) Prove that f '(x) is continuous for x # 0 and that lim f '(x) = f '(0) = 0,

x-to

... . so that f '(x) is continuous for all x. c) Prove that f (")(x) is continuous for all x and f (")(O) = 0. d) Graph the function f (x). 7. Use the remainder formula to estimate the error in the following computations: a) e = 1 + 1 +

+ +& +4 1

[assume that e < 3 is known)

1

c ) l o g ; = ; - ~ + ~ 8. Let f (x) satisfy the conditions stated in Theorem 41. Let f '(a) = f " ( a ) = . . . = f (")(a) = 0, but f ("+')(a)# 0. Show that f (x) has a maximum, minimum, or horizontal inflection point at x = a according to whether the function f("+')(a)(x - a)"+' has a maximum, minimum, or inflection point for x = a. (This gives another proof of the rule deduced in Section 2.19.) 9. (Derivatives and differences) In numerical analysis, one uses approximations for derivatives. For the first derivative of f (x) at x, one chooses a small positive h and uses

For the second derivative, one uses

With the aid of Taylor's formula with remainder (assuming sufficient differentiability), establish the following: a) g l ( x , h ) - f'(x)haslimitOash + Oand

hasafinitelimitash-r 0.

b) g2(x, h) - f '(x) has limit 0 as h -+0 and 8-

has a finite limit as h -r 0.

g3(x. h) - f "(x) has limit 0 as h -r 0 and

has a finite limit as h -r 0.

C)

Remark We describe these results by saying that the error in approximating fl(x) by gl(x, h) or ftl(x) by g3(x, h) is of the order h, while the error in approximating f1(x) by gz(x, h) is of the order h2. (The higher the power of h, the better the approximation for small h.)

0

438

Advanced Calculus, Fifth Edition

*6.19 SEQUENCES AND SERIES OF COMPLEX NUMBERS We assume familiarity with complex numbers, written typically as z = x i2 = - 1. The absolute value of z is defined as

'

",.'

+iy, where

and hence lzl equals the magnitude of the vector (x, y) of v2 or the distance from z to the origin in the xy-plane. Since addition and subtraction of complex numbers is the same as for vectors in v2,Jzl- z21 can be interpreted as the distance from zl to 22. These concepts are illustrated in Fig. 6.12. Sequences of complex numbers are defined as for real numbers. The following are examples:

A sequence 2, is said to converge to 20: lim zn = 20

n+m

if, given 6 > 0, an integer N can be found such that

,zI -zOI

<E

forn > N.

This resembles the definition for real numbers. The inequality states that z, is within distance r of zo or that z, is within the circle of radius r about zo; this is illustrated in Fig. 6.13. If a sequence z, does not converge, it is said to diverge.

a

Figure 6.12

Complex numbers.

Figure 6.13

Sequence converging to zo.

Chapter 6 Infinite Series

439 %

+

iy, ( n = 1,2, . . .) be a sequence of complex THEOREM 46 Let z , = x, numbers. If this sequence converges to zo = xo iyo, then

+

lim x, = xo,

n+w

lim y, = yo.

n+w

Conversely, if x, converges to xo and y, converges to yo, then

This theorem shows that the convergence of sequences of complex numbers can be referred back to that of real numbers simply by studying real and imaginary parts. To prove the theorem, we remark that, if lzn - zol < r , then (x,, y,) is within the circle of radius r about (xo,yo), so that necessarily Thus convergence of z , to zo implies convergence of x, to xo and y, to yo. Conversely, if x, converges to xo and y, converges to yo, then for given r, N can be chosen so large that 1 1 ly, - yol c -6 for n > N. Ix, - xol < -E, 2 2 These inequalities force ( x , , y,) to lie within a square with center (xo,yo) and side r ; hence (x,, yn) must lie within the circle of radius r about (xo, yo), so that lz, - zoI < r for n > N. Accordingly, z , must converge to zo.

THEOREM 47 (Cauchy Criterion) A sequence z , of complex numbers converges if and only if for each r > 0 and N can be found such that

This is proved by refemng the convergence of z , back to that of the two real sequences x,, y, and then applying the Cauchy criterion (Theorem 6) to the real sequences. An in$nite series of complex numbers is defined as for real numbers:

The series is said to converge or diverge according to whether the nth partial sums, form a convergent or divergent sequence. The sum of the series is then the limit S = lim S, n+w

when the limit exists.

440

Advanced Calculus, Fifth Edition On the basis of Theorem 46, one can at once assert the following theorem: THEOREM 48

If zn = xn

+ iy,, then the series

converges and has sum S = A

+ Bi if and only if

Thus the convergence of series of complex numbers is also referred back to that for real numbers. This can also be done in a second way.

THEOREM 49 If

CEl JznI converges, then xElzn converges.

a

In words: If a complex series is absolutely convergent, then it is convergent. The proof is the same as that of the corresponding theorem for real numbers (Theorem 11 of Section 6.6). On the basis of Theorems 48 and 49, one can now obtain tests for convergence and divergence of series of complex numbers. In particular, the following rules hold: the nth term test (Theorem 10); the Cauchy criterion (Theorem 9); the comparison test for convergence (Theorem 12); the ratio test (Theorems 17 and 20); the root test (Theorems 19,21, 21(a)). The rule for addition or subtraction of convergent series (Theorem 8) and the product rule (Theorem 29) also hold for complex series. The estimates of remainders (Theorems 22,23,24,25) can also be used for complex series. EXAMPLE 1

The series

is absolutely convergent since the series of absolute values is the real series

which converges (for example, by the ratio test).

EXAMPLE 2 The series i

i2

i3

in

2

3

n

-+-+-+...+-+... 1

= n=l

.

-

=

.-

Chapter 6 Infinite Series is not absolutely convergent since C l l n diverges. However, the series of real parts is

and the series of imaginary parts is

1 1 1+0--+o+-+-.. . 3 5 If the zeros are disregarded, these are convergent alternating series. Hence converges.

in/n

The general theory of functions of a complex variable is developed in Chapter 8. Here we consider briefly the functions zn, where n is a positive integer or 0 , and the corresponding power series: 00

'&zn

= co + c l z + c 2 z 2

+ - - .+c,zn + .

. a .

(6.56)

n=O

Since each term is defined for all z, the series may converge for some or all z. The basic theorem on power series (Theorem 35) can be restated and proved just as for real numbers, and one concludes that each power series (6.56) has a radius of convergence r* such that the series converges when lzl < r* and diverges when > r*. The use of the term radius of convergence now receives its justification; for the power series (6.56) converges within the circle with center 0 and radius r * .This is illustrated in Fig. 6.14. It is of special interest that the power series (6.46), (6.47), (6.48) for e x ,sinx, and cos x continue to converge when x is replaced by an arbitrary complex number z. One can then use the equations

Figure 6.14

Radius of convergence of power series.

441

442

Advanced Calculus, Fifth Edition to dejne these functions for complex z. From these series, one derives the Euler identity (Problem 4 below): e i ~= cosy

+ i sin y,

(6.60) .?I ?c

or the more general relation:

This last equation can also be used as a dejnition of eZfor complex z, the series expression being then a consequence of this definition. This is shown in Chapter 8.

PROBLEMS 1. Evaluate the limits: b) limn-+w ( 1 + i)n3 - 2in + 3

in

a)

in'

-1

2. Test for absolute convergence and for convergence: a) (!+)+(w)2+.-+(T)

ni"

m

Cn=l ,,2+1

l+i n

+...

b)

c

C z lnin Cn=l 1 00

d) 3. Prove that the series (6.57), (6.58), and (6.59) converge for all z. 4. a) Prove the Euler identity (6.60) from the series definition of eZ,cos z, sin z. b) Prove (6.61) from the series definitions of eZ,cosz, sinz. 5. Use the series expressions (6.57), (6.58), (6.59) to prove the identities C)

erz + e - l z

a) cosz = 7 C) e Z ~+22 = ezt . e ZZ

e) cos (-Z) = cos z g) cos 22 = cos2z - sin2z

erz

- e-'z

b) sin^=.^ d) sin (-Z) = - sin z f ) sin2z cos2z = 1 h) sin 22 = 2 sin z cos z

+

6. Show that the following series converge for lzl < 1 and diverge for lzl > 1: 2

a) z + % + . . . + $ + . . .

b) 1 + z + z 2 + . . . + z n + . . -

*6.20 SEQUENCES AND SERIES OF FUNCTIONS OF SEVERAL VARIABLES The notions of sequence and series of functions extend at once to functions of several variables. Thus

is a series of functions of the two variables x and y. The notion of uniform convergence also extends at once, as well as the M-test and the properties described in Section 6.14. t

:

i

Chapter 6 Infinite Series

Figure 6.15

Convergence set for (6.62).

:1

c

One can in particular considerpower series in several variables. For two variables x , y, such a power series is a series

where fn(x, Y) = cn,oxn

+ cn,lxn-'y + ... + cn,n-lxy + c,,nyn, n-1

(6.64)

the c's being constants. Thus fn is a homogeneous polynomial of degree n in x and y . The series (6.62)is an example of this, with f2, = x n y n and fo = 0, fl = f3 = ... = 0. This series also illustrates the fact that the values (x, y ) for which a power series in x and y converges form a more complicated set than the convergence interval for series in x alone. For (6.62) the series converges when lxy 1 < 1; this region is pictured in Fig. 6.15. In general, the convergence region can be quite complicated. If a function F ( x , y) can be represented by such apower series in a neighborhood of the origin,

a

then a term-by-term differentiation (which can be justified) shows that

.t

ei&

443

444

Advanced Calculus, Fifth Edition In general one finds

all derivatives being evaluated at (0,O). A series C fn(x, y), in which the fn are given by (6.65), is known as aTaylor series inx and y, about (0, O), and the function F ( x , y) that it represents is termed analytic in the corresponding region. The expansion about a general point (XI,yl) is obtained by a translation of origin:

all derivatives being evaluated at (xl, yl). The general term of the series (6.66) can be interpreted in terms of an nth diflerential dnF of the function F(x, y):

where the (;) are the binomial coefficients.' To indicate the dependence of d n Fon XI,yl and the differences x - x l , y - yl, we write: d n F = dnF(xl,yl; x - X I ,y - yl). When n = 1 and x

- xl = dx, y - yl = dy, one finds:

this is the familiar expression for the first differential. The series (6.66) can now be written more concisely:

Chapter 6 Infinite Series The theory of analytic functions of several variables is again best studied with the aid of complex numbers. As with one variable, the familiar functions are in general analytic. Thus

the series converging for all x and y . The subject of analytic functions of several variables has not been studied intensively until recent times, and most books on the subject are very advanced. In 0.D. Kellogg's Foundations of Potential Theory (Berlin: Springer, 1967) a brief treatment is given on pages 135-140. An introduction to the topic is also given in the last chapter of the book Introduction to Analytic Functions by W. Kaplan (Reading, Mass.: Addison-Wesley, 1966).

*6.21 TAYLOR'S FORMULA FOR FUNCTIONS ' OF SEVERAL VARIABLES

'

There is a Taylor formula with remainder for functions of several variables:

The point ( x * , y * ) lies between (XI,y1) and ( x , y ) on the line segment joining these points, as in Fig. 6.16. For n = 1 the formula becomes

This is known as the Mean Value theorem for functions of two variables. To prove (6.68), one writes:

Figure 6.16

Taylor's formula for F(x, y).

445

446 .

Advanced Calculus, Fifth Edition Thus x and y are considered as fixed, and 4 depends only on t . By the Mean Value theorem for 4, ,zt d

1

=

o

+t

,r2-:

* ,

o < t* < 1.

If t is replaced by t*, one obtains (6.68). The general formula (6.67) is proved in the same way, on the basis of Taylor's formula for 4:

where 0 < t* < 1. For one finds by induction that

4'"'(t) = dnF[xl+ t ( x - X I ) yl ,

+ t ( y - y l ) ; x - X I , y - yl].

(6.69)

The validity of (6.67) is ensured if F ( x , y) has continuous derivatives through the ( n 1)st order in a domain D containing the line segment joining ( x , y ) to

+

(XI9 ~ 1 ) .

Taylor's series or Taylor's formula can be used to study the nature of a function near a particular point. As was remarked earlier, the linear terms give dF, the best "linear approximation" to F ( x , y ) - F ( x l ,y l ) . If dF = 0, the quadratic terms d 2 ~ / 2 ! become of importance. In particular, if the quadratic expression

is positive, except for x = x l , y = y l , then F ( x , y ) has a minimum at ( x t ,yl). Pursuing this further, we rediscover the criteria for maxima and minima developed in Section 2.19. Remark. The first differential dF of F ( x , y ) can be considered as an "invariant differential form." By this we mean that, under a differentiable substitution x = x ( u , v ) , y = y(u, v ) , the differential of the composite function F ( x ( u , v ) , y(u, v ) ) equals that of F at corresponding points:

aF dF(x(u,v ) , Y ( U v, ) ) = ax dx

aF

+dy, ay

whereontherightx = x ( u , v ) , y = y(u, v ) , d x = ( a x / a u ) d u+ ( a x / a v ) d v , d y = ( a y / a u ) d u ( a y / a v ) d v . The assertion follows at once by the chain rule (as in the theorem concerning Eq. (2.36) in Section 2.8). For the second and higher order differentials, the analogous invariance fails in general. However, at a point ( x l ,y l ) at which dF = 0, d 2 = ~ 0, . . . , dn-'F = 0, the differential dnF is invariant, so that under the appropriately differentiable substitution x = x ( u , v ) , y = y ( u , v ) , d n F ( x ( u ,v ) , y ( u , v ) ) is calculated from dnF(x, y ) by expressing x , y, d x , d y in terms of u , v , d u , d v and expanding by algebra. For example, from

+

447

Chapter 6 Infinite Series d 2 =~Fxx(x,y)dx2

+ 2Fxy(x,y)dx dy +'F,,(x,

y)dy2 one obtains

-r

d 2 ~ ( x ( uv), , y(u, v ) ) = Fxx(x(u,v), y(u, v ) ) at each point (ul, vl) at which x = xl, y = yl and Fx(xl, yl) = 0,Fy(xl, yl) = 0. (See Problem 6 below.) The invariance of d2F at a critical point explains its importance for maxima and minima as in the preceding paragraph. The concept also occurs in discussion of the secondfindamental form on a surface (Problem 23(d) following Section 3.1 1). See also Sections 5.19 and 5.20.

PROBLEMS 1. Expand in power series, stating the region of convergence: a) e ~ 2 - ~ 2 b) sin(xy) c)

1

I-n-y

1

d) 1 - x - y - z

2. Prove Eq. (6.69) by induction. 3. Prove that if a power series.

converges at (xo,yo), then it converges at every point (Axo,hyo), for 1 Al c 1. 4. Evaluate J; 1; sin (xy) dx dy with the aid of power series. 5. As in Problem 9 following Section 6.18, use Taylor series or the remainder formula to analyze the following difference approximations to partial derivative expressions: a) g , ( x , h ) = f ( x + h , y + h ) + f ( x + h , y - h ) h2

+

f(x-h,y+h)+f(x-h.y-h)-4f(x3y) h2

as approximation to v2f at (x, y); b) g2(x,h)= f ( ~ + h . ~ + h ) - f ( x + h , y - h ) - f ( x - h , y + h ) + f ( x - h , ~ - h ) 4h2 4i as approximation to a2f/ax ay at (x, y). 6. Prove the validity of Eq. (6.71) at (ul, vl) under the hypotheses stated. [Hint: Use the rules of Section 2.16.1

In Section 4.1, improper integrals are defined. In this section we show that there is a close relationship between improper integrals and infinite series. The relationship is suggested by the integral test of Section 6.6. It leads us here to some valuable tests for convergence of improper integrals. Attention will be confined to integrals from a to oo, but the methods are applicable to other forms of improper integrals. Let us consider first, as an example, the integral

448

Advanced Calculus, Fifth Edition This would be termed convergent if the limit

exists. However, since the integrand l l x is positive, the integral from 1 to b must increase as b increases. From this it follows that either the above limit is +oo (that is, there is no limit) and the integral diverges or else the limit is a finite number I. Now, in order to determine which of the two cases holds it is clearly sufficient to let b approach oa through integral values n , that is, to consider the limit of a sequence:

But one can write

.

In other words, the integral exists precisely when the series

converges and the two have the same value. This is suggested in Fig. 6.17. In this particular case,

n+l = log -, n

I

and the series is

One easily verifies that both series and integral diverge. We now formulate the relationships in general terms.

Figure 6.17

Improper integral as a series.

449

Chapter 6 Infinite Series

THEOREM 50 Let f (x) be continuous and let f (x) 2 0 for a 5 x

-Z ao.The2

the integral

lW f(x)dx

converges and has value I if and only if the series

converges and has sum I. If f (x) changes sign infinitely often, then convergence of the integral will certainly imply that of the series; however, the converse need not hold, as the example

liD

sin 2irx dx

(6.72)

shows (Problem 2 following Section 6.25). The proper connection between series and integrals in this case is given by the following theorem. THEOREM 51 Let f (x) be continuous for a 5 x c oo. Let f (x) 2 0 for a = bo 5 x i bl, f (x) 5 0 for bl 5 x 5 b2 and, in general, (-1)" f (x) 2 0 for b, 5 x _( bn+l,where b, is a monotone sequence such that

lim b, = oo.

n-+m

Then the integral

converges and has value I if and only if the alternating series i

converges and has sum I. Proof.

If the integral converges to I , then as before, lim

n-tw

lb"

f(x)dx = lim(al n+ca

+

so that the series converges and has sum I. Conversely, let the series converge to I. Let large that 1 la~+...+a,-I1 < - r 2

and

+a,) = 1,

6

> 0 be given and choose N so"-

1 la,l<-E 2

forn~N.

450

Advanced Calculus, Fifth Edition The latter condition can be satisfied by the nth term test. If xl > bN,then-b, 5 xl 5 b,+l for some n 2 N. Hence

and

the last step is justified, since f (x) does not change sign between b, and b,+l. Hence

It follows that

lili

XI

f(x)dx = I .

COROLLARY Let f (x) be continuous for a 5 x < oo, let f (x) decrease as x increases, and let lim,,, f (x) = 0.Then the integrals

Jm

Jwf (x) cos x dx

(x)sinx dx,

converge. Proof. We consider the sine integral, the cosine integral being similar. Under the assumptions made, the alternating series of Theorem 5 1 is, except perhaps for the first term, of the form (n+l)rr

Fan,

an=ln

,,

f (x)sinx dx.

"

:.,

,z.q*

?*f-$.

-- -

n=k

Since f (x) decreases as x increases, la, 1 is decreasing; since f (x) has limit 0 as x -+ oo, a, converges to 0. The alternating series test (Theorem 18) then guarantees convergence. EXAMPLES The integrals

sin x

lw

dx,

all exist, by virtue of the preceding corollary.

dx,

l m y d x

451

Chapter 6 Infinite Series

THEOREM 52 (Cauchy Criterion) Let f (x) be continuous for a 5 x < oo. The integral

JW

f (x)dx

exists if and only if for each r > 0 a number B can be found such that I['f(~)d~l < s

for B < p < q.

Proof: If the integral converges to I , then for given 6 > 0 a B can be found such that

Hence for B < p < q ,

and

Conversely, let the condition hold. Then the Cauchy criterion of Theorem 6 applies to the sequence

hence this sequence converges to a number I. Let E > 0 be given and let a corresponding number B be chosen as in the theorem; let N be an integer larger than B and such that IS, - 11 < r forn 2 N. Forx, > N,

It follows that lim X1'cQ

lx'

f(x)dx = I.

21

452

Advanced Calculus, Fifth Edition Remark.

It follows from Theorem 52 that for a convergent integral

Xrn*f ( x )d x ,

for every fixed k. However, it is not necessary that f ( x ) itself approach 0.This is illustrated by the integral (Problem 1 following Section 6.25):

lW sin x2 d x .

THEOREM 53 Let f ( x )be continuous for a 5 x c oo. 1f Jam If (x)l d x converges, then

converges, and

In words: An absolutely convergent improper integral is convergent. Proof. Since If 1 is a continuous function of f , If (x)l is a continuous function of a continuous function and must hence be continuous. Now

Hence if the Cauchy criterion holds for the integral of If 1, it must also hold for the integral of f itself; thus the convergence of j" If 1 d x implies that of J f d x . Furthermore,

This is true for every b; hence in the limit,

THEOREM 54 (Comparison Test) Let f ( x ) and g ( x ) be continuous for a 5 x c oo. If 0 5 If (x)l 5 g ( x ) and Jam g ( x ) d x converges, then

is absolutely convergent, and

6

Chapter 6 Infinite Series

diverges, then

diverges.

*:i0'2

The proof is just like that for series and need not be repeated. The preceding discussion has been confined to integrals from a to oo.A similar one holds for integrals from -oo to b and for integrals from a to b, which are improper at an endpoint. Thus the convergence of integrals

where f is unbounded near 0 and continuous for 0 < x I1 , is related to the convergence of series

where bn is a monotone sequence converging to 0.

The analogue of a series of functions

is an improper integral

lm f(t,x)df;

thus the variable t replaces the index n. Both of these, when convergent, define a function F ( x ) . Because of the close relationship between series and improper integrals demonstrated in the preceding section, we can expect the discussion of functions defined by integrals to parallel that of Sections 6.11 to 6.14 for functions defined by series. One defines the improper integral to be uniformly convergent to F ( x ) for a given range of x , if, given E > 0, a number B can be found such that

where B is independent of x.

--

.

453

Advanced Calculus, Fifth Edition THEOREM 55 (M-Test for Integrals) Let M ( t ) be continuous for a 5 t < w; let f ( t ,x ) be continuous in t for a 5 t < oo for each x of a set E. If

If

( t ,x)l 5 M ( t )

for x in E and

converges, then

is uniformly and absolutely convergent for x in E. THEOREM 56 and

If f ( t ,x ) is continuous in t and x for a 5 t < oo,c 5 x 5 d ,

is uniformly convergent for c 5 x 5 d , then the function F ( x ) defined by this integral is continuous for c 5 x 5 d . THEOREM 57 and

If f ( t ,x ) is continuous in t and x for a 5 t < w, c 5 x 5 d ,

is uniformly convergent to F ( x ) for c 5 x 5 d , then

ld

F ( x )d x =

lw ld

f ( t , x ) d x dl.

THEOREM 58 If f ( t ,x ) is continuous in t and x and has a derivative af l a x , which is continuous in t and x for a 5 t < oo and c 5 x 5 d , and the integrals

converge, the second one uniformly, for c 5 x 5 d , then

has a continuous derivative for c 5 x 5 d , and F1(x)=

lm \.i z ( t ,x ) d t .

These theorems are proved exactly as for series.

.-

-,-*+,

a

,. r J i i '

Chapter 6 Infinite Series

EXAMPLE The integral

is uniformly convergent for x

> 1, since

0 5 e-Xt25 e-'2 = M(t) for x s 1, and the integral

exists, as a comparison with

&. d

reveals. Hence (6.74) defines a function F(x). Theorem 56 shows that F(x) is continuous for x > 1. The integral

is also uniformly convergent, since

for x 2 1 and t sufficiently large. Hence

"hm

e-xr2

dt = -

I

"

i t'e-"2

dt,

x 2 1.

dx

As in the preceding section, the theory extends to improper integrals over a finite interval without essential change.

An integral from -oo to oo would normally be decomposed into integrals from -00 to 0 and from 0 to oo,as in Section 4.1. If the last two integrals converge, the given integral converges; otherwise, it diverges. When this process leads to divergence, one may be able to salvage the integral by the following procedure, which has important applications to Laplace and Fourier transforms. Let f be continuous for -oo < x < oo (or, more generally, let f be such that f has an integral over each finite interval). Then the Cauchy principal value (briefly, principal value) of the integral of f from -oo to oo is the limit

if this limit exists. When it does, one denotes the value by

456

Advanced Calculus, Fifth Edition One sees at once that if

I-",

lirn

a+co

f ( x )d x exists in the usual sense, then

la

la 0

f ( x )d x and

:li

f ( x )d x

both exist, and hence the principal value exists and equals the usual improper integral. However, the principal value may exist even though the usual value does not. For example,

1 1 + ( a - 1 P =, = lirn - log a-.a,2 1 + ( ~ + 1 ) ~

Here the integral from 0 to oo is +oo, from -oo to 0 is -oo. There is in effect a cancellation of the two infinities. This is related to a certain symmetry of the graph of the function. The concept of principal value can be extended to the integral of a function f from a to b, where f is discontinuous only at c, with a < c < b. Here one defines:

f ( x )d x = lim €'O+

[LC-'

f ( x )d x

+Jb

( x )d x ] .

c+c

For example,

Here the integrals from 0 to 1 and from 1 to 3 are +oo and -oo,respectively, and again there is a cancellation of the two infinities, made clear in the evaluation of the last limit. It is easily seen, as previously, that whenever the integral of f from a to b exists, as usual, as an improper integral, the principal value exists and equals the previous value; the example just given shows that the principal value can exist even when the usual improper integral does not exist. Of course, the principal value can also fail to exist, as the following example shows:

d x = lim c+O+

= a+O+ Iim

[[y f [(: +

dx

1)

+

1

; id x ]

+ ( - 1 + f)] =

Here there is no cancellation because the function is always positive. If f is continuous for -00 c x c oo except at c, then one can use a principal value both for c and for large x as follows:

(P)

/

c-E

03

-w

f ( x )d x = lirn lirn

f(x)dx

+Sa

f(x)dx].

c+c

This procedure can be adapted to the case of several discontinuities c l , . . . , c,.

.

nr~.

Chapter 6 Infinite Series

*6.25 LAPLACE TRANSFORMATION I?-FUNCTION AND B-FUNCTION In view of the analogy between infinite series and improper integrals it is natural to seek an improper integral corresponding to a power series. If we write the power series as

then a natural analogue is the improper integral

lw f (r)xt d t .

(6.76)

Except for a minor change in notation, this is the Laplace transform of the function f (t). More precisely, the Laplace transform of f (t) is the integral

(6.77) is obtained from (6.76) by replacing x by e-S. Just as (6.75) defines a function G(x) within the interval of convergence, the integral (6.77) defines a function F(s), for those values of s for which the integral converges: F(r) =

Lrn

f (t)e-Stdt.

In fact, the integral (6.76) can be shown to have a "radius of convergence" r* such that (6.76) converges for 0 5 x < r*; it is necessary to restrict to positive x, since x' would be imaginary for x negative and t = etc. Accordingly, the integral (6.78) converges and defines a function F(s) for 0 < e-* < r*, that is, for s > log (llr*). Integrals of form (6.78) have proved to be exceedingly useful in the theory of ordinary and partial differential equations. The integral can be regarded as an "operator" 2'transforming the function f (t) into the function F ( s ) , and one writes

i,i,

This explains the word transform and also the use of the term operational method in connection with the applications to differential equations. (See Section 7.19 and the book by Kaplan listed at the end of Chapter 7.) An important particular Laplace transform is:

in which f (t) = tk;the parameter k must be greater than - 1, to avoid divergence of the integral at t = 0. With k so restricted, the integral converges for s > 0 (Problem 1). When k is 0 or a positive integer, the integral is easily evaluated:

457

Advanced Calculus, Fifth Edition In general, an integration by parts (Problem 10) shows that

Accordingly, we can apply induction to conclude that

For s = 1, (6.81) gives k! =

L

00

tke-' d t .

This suggests a method of generalizing the factorial; that is, we could use Eq. (6.82) to define k! for k an arbitrary real number greater than - 1. It is customary to denote this generalized factorial by r ( k f 1); the Gamma function r(k) is then defined by the equation

when k is a positive integer or 0,

The general integral (6.79) is expressible in terms of the Gamma function (Problem 14):

Equation (6.80) then states that

that is, This is thefunctional equation of the Gammafunction. The functional equation can be used to define r(k) for negative k; thus we write

4).

3).

in order to define r ( r(. . . in terms of the known value of r(f ). This procedure fails only for k = - 1, -2, . . . . In fact, we can show (Problem 13) that lim

k+O+ 1

r (k) = +oo,

-

Chapter 6 Infinite Series

Figure 6.18

y = r(x).

and if the Gamma function is extended to negative nonintegral k as before, lim Ir(k)l = +oo

k-+ -n

for every negative integer -n. These properties are indicated in Fig. 6.18. The Gamma function has other important properties: 1 = keYklim

[

( + :) .. . + )!

(1 + k ) 1 r(k) n+m here y is the Euler-Mascheroni constant (Problem 15):

r(k) = kk-4e-k&ee(k)/12k,

e-k-:--!];

k > 0,

(6.87)

(6.89)

where 8(k) denotes a function of k such that 0 < 8(k)< 1. Proof of (6.87) and (6.89) and other properties of the Gamma function are given in Chapter 12 of the book of Whittaker and Watson listed at the end of this chapter. From (6.89), one can prove (Problem 16) that +

lirn

k

+

r(k kk+f ~ a

+ 1) e - k

= 1.

When k is an integer, this gives the Stirling approximation to k!:

Fork = 10the left-hand side is 3.629 x lo6 whereas the right-hand side is 3.599 x lo6. The Beta function B(p, q) is defined by the equation

This can be expressed in terms of the Gamma function (Problem 17):

459

Advanced Calculus, Fifth Edition

PROBLEMS 1. Determine convergence or divergence of the following improper integrals: em+

a) C)

b, d)

7dx

jp" 9

dx

JI

[IO~X)~

JOo0

tke-st d t , k > -1

sinx2 d x (Hint: Let u = xZ.)

e)

2. Prove that if f (x) = sin 2nx, then the series

Jr

converges, but f (x) dx diverges. 3. Prove the ratio test for integrals: If f (x) is continuous for a 5 x < oo and

ST

then f (x) dx is absolutely convergent. 4. Apply the ratio test of Problem 3 to prove convergence of the integrals:

JF 5

b) $ dx a) dx 5. Prove the root test for integrals: If f (x) is continuous for a 5 x < oo and lim 1 f(x)l'lx = k < 1,

17

X"'

then S,bO f (x) dx is absolutely convergent. 6. Apply the root test of Problem 5 to prove convergence of the integrals:

a)

J,OO e-l2 dx

00

dx

b, S2 (10gx)~ 7. Prove that the following integrals are uniformly convergent for 0 5 x 5 1: dt b) SpO* X a) ( x 2 + 12); 8. a) Prove that for every choice of xl > 0 the integral p e - x t z d t , n > 0,

iy

dt

is uniformly convergent for x 2 X I . b) Use the known result that

I"

e-X2dx =

-21 f i

(Problem 1 following Section 4.8) to prove that

C)

Use the results of parts (a) and (b) and Theorem 58 to show that for n = 1,2, . . . , and x > 0,

9. a) Prove that if n > 0, the integrals

1"

tne-t2

cos (tx) a t ,

are un~formlyconvergent for all x .

I"

pe-t2

sin . (tx) d t

Chapter 6 Infinite Series b) Let F(x) =

lm

e-" cos (rx) dr.

Use integration by parts (cf. Problem 10) to show that F1(x) = , $ x ~ ( x )From . this, deduce that d log F(x) = - i x d x and hence that F(x) = ce-ax . Let x = 0 and use Problem 8 to find c and thus prove that

10. a) Prove that if u(x), ul(x), v(x), vf(x) are continuous for a 5 x < oo and limX,,[u(x)v(x)]

I"

exists, then

u(x)vf(x)dx = lim [u(x)v(x)] - u(a)v(a) -

" ul(x)v(x) dx;

X"'

that is, if one of the two improper integrals converges, then the other converges and the equation holds. b) Use the result of part (a) to justify the derivation of Eq. (6.80). C) Prove Abel's formula:

and hence obtain the analogue of part (a) for infinite series:

that is, if limn+,(unvn+l) exists, then convergence of one of the two series implies convergence of the other and validity of the equation. This "integration by parts" for series provides valuable tests for convergence; cf. Chapter X of the book of Knopp listed at the end of the chapter. 11. Prove the following: a) i f f is continuous for -oo < x < oo and f is odd, then (P):-f f (x)dx = 0; b) iff is continuous for -a I:x 5 a , except at x = 0, and f is odd, then (P)a:f f (x) dx = 0. 12. For each of the following integrals, obtain the principal value, if it exists; if it does, determine whether the usual value also exists [Hint: Note the result of Problem 11.I: a)

:-I -$dx x +1

b)

l-:

c) :-f

& d xi

4 f) L-2

e) j!, $ d x 9) : -j

h)

$ dx

"

f-"

sin x d x 1

dx

dx

13. a) Prove from Eq. (6.83) that r ( k ) is continuous and positive for k > 0. b) Prove from Eq. (6.86) that

,

,

lim r ( k ) = +oo; k+O+ C) Prove that if Eq. (6.86) is used to define r(k) for k negative but nonintegral, then lim Ir(k)l = +GO

k+-n

for n = 0, 1,2..

..

461

462

Advanced Calculus, Fifth Edition 14. Prove (6.85). 15. a) Prove by the integral test that the series (6.88) defining y converges. b) Use Theorem 23 to prove that y = 0.6 to one significant figure. 16. Prove (6.90) from (6.89). 17. Prove the identity (6.93). [Hint: Show that

t:

and hence, as in Problem 1 following Section 4.8, that

The first factor on the right is T ( p + q); the second reduces to B ( p , q ) if one sets x = sin28.1

18. Verify that the following functions F ( s ) are Laplace transforms of the functions f ( t ) given: a) F ( s ) = = ,1 s > ' k ; f ( t ) = e k t ; b) F ( s ) = - , k s

+k

s > O ; f(t)=sinkt

For simplicity we confine attention to double integrals and consider the case of f (x, y) continuous but unbounded on a set R consisting of a bounded closed region minus one point (xo, yo) interior to the region. We can further take R to be a circular region with center (xo, yo) (excluded) and radius a since we can always consider the integral over R to be an improper one over such a circular region plus a "proper" integral over the rest of R . The improper integral of f over R is said to exist and have value c (Section 4.8) if for every E > 0 we can find a bounded closed region R, contained in R such that for every bounded closed region R2 contained in R and containing R1 we have

Our purpose is to show that in certain cases this limit process'can be simplified by considering only

where Eh is the part of R outside a circular neighborhood of (xo, yo) of radius h.

Chapter 6 Infinite Series By analogy with the definitions of Section 6.24 we call (6.94)the principal value of the double integral of f over R. We point out that since R is a circular region, we can write the limit (6.94)as

..

lim h+O+

la h

1 2 * f r dr dB.

(6.95)

+

+

Here we have introduced polar coordinates at (xo,yo):x = xo r cos 9, y = yo r sin 8, so that f is continuous in r, 8 for all 0 and 0 c r _< a. By Theorem M of Section 4.1 1 the inner integral defines a function

continuous for 0 < r 5 a, and we are really considering an improper integral

la

rg(r)dr = lim h+O+

sa h

rg(r)dr.

(6.96)

We can parallel the discussion of Section 6.22 for such integrals. For example, if a > 1 and g(x) 2 0 for 0 5 x 5 a , then the integral exists if and only if the monotone increasing sequence ilc)

is bounded. Also, in general, the integral exists if and only if there is a 6 > 0 such that

(Cauchy criterion). The theorem on absolute convergence and the comparison test also have their counterparts.

+ -

THEOREM 59 Let R be the region 0 < (x - X O ) ~ (y yo)' 5 a*, where a is a positive constant. Let f be continuous and unbounded on R and let the principal value of the double integral of f over R be absolutely convergent, so that in the notation given earlier,

lim

h+O

lalLr1 h

f )r d9 dr

exists. Then the improper double integral of f over R exists. Proof. We introduce g(r) as before for f and also gl(r) for If 1, so that

463

464

Advanced Calculus, Fifth Edition The existence of the limit (6.98) is equivalent to existence of the improper integral

Hence by the Cauchy condition we can choose 6 so small that for given r > 0,

1:'

E

rgl(r)dr c - for 0 c h2 c hl < 8 . 2

(6.101)

Now 2n

Ig(r)I 5

C

If(x~+...,yo+...)ldO = g ~ ( r ) .

Hence existence of the integral (6.100) implies, by comparison, that the integral (6.96) exists, so that the principal value (6.94) for f exists. We let c be this value. Given r > 0, we now choose S as before but also so small that

for 0 < h < 6. We choose ho, h, so that 0 < h l < ho < 6 and let Ri be the annular region Eh, for i = 0, 1. Then

Now let R2be any bounded closed region containing R1 and contained in R. Then Rz can be considered as the union of the annular region Ro and a region R3 contained in the circular region (x - X O ) ~(y - yo)25 hi but not containing (xo,yo) (Fig. 6.19).

+

Figure 6.19

Proof of Theorem 59.

Chapter 6 Infinite Series Therefore R3 lies in some annulus h2

< r < ho with hz > 0.Hence

Thus

as was to be proved.

COROLLARY In Theorem 59 let f ( x , y ) = p ( x , y ) log ( l / r ) or p ( x , y)r-P for 0 < p < 2, where p ( x , y ) is bounded in R. Then the improper double integral o f f over R exists. a The proof is left as an exercise (Problem 2 below).

PROBLEMS

+

+

1. Let f ( x , y ) = (x y ) / r P , where r = ( x 2 y 2 ) ' / 2 and p = const > 0. Show that the principal value of the double integral of f over R: 0 < r 5 1 exists, but the double integral itself exists only for p < 3. 2. Prove the corollary to Theorem 59. 3. a) Let R be the solid sphere of radius a and center (xo,yo, zo) with the center deleted. Let f ( x , y , z ) be continuous and unbounded on R. Define the principal value of the triple integral of f over R by analogy with (6.94) and state and prove the analogue of Theorem 59 for f ( x , y , z ) . b) On the basis of the theorem of part (a) deduce that if g(x, y , z ) is bounded and continuous over R, then the improper integral

exists if 0 < p < 3. Suggested References Courant, Richard J., and Fritz John, Introduction to Calculus andAnalysis, vol. 1, New York: Interscience, 1965 and vol. 2, New York: John Wiley & Sons, 1965. Green, J. A., Sequences and Series. Glencoe, 111.: The Free Press, 1958. Hardy, G. H. Pure Mathematics, 10th ed. Cambridge: Cambridge University Press, 1952. Henrici, P., Applied and Computational Complex Analysis, vol. I. New York: John Wiley & Sons, 1991.

465

466

Advanced Calculus, Fifth Edition Knopp, K., Theory and Application of InJnite Series, trans. by Miss R. C. Young. London and Glasgow: Blackie and Son, 195 1. Rudin, W., Principles of Mathematics Analysis, 3rd ed. New York: McGraw-Hill, 1976. Whittaker, E. T.,and G. N. Watson, Modem Analysis, 4th ed. Cambridge: Cambridge University Press, 1950. Widder, D. V., The Luplace Transform.Princeton: Princeton University Press, 1946.

Fourier Series and Orthogonal Functions

A trigonometric series is a series of form 1

where the coefficients a, and b, are constants. If these constants satisfy certain conditions, to be specified in Section 7.2, then the series is called a Fourier series. Almost all trigonometric series encountered in physical problems are Fourier series. Each term in (7.1) has the property of repeating itself in intervals of 2n: COS(X+~~)=COSX s i,n ( x + 2 n ) = s i n x ,

...,

cos [n(x + 2n)] = cos (nx + 2nn) = cos nx, . . . . It follows that if (7.1) converges for all x, then its sum f (x) must also have this property:

f (x + 2n) = f (x).

(7.2)

We say: f (x) has period 275. In general, a function f (x) such that

for all x is said to be periodic and have period p. It should be noted that cos 2x has, in addition to the period 2n, the period n and, in general, cos nx and sin nx have the periods 2nln. However, 2n is the smallest period shared by all terms of the series.

Advanced Calculus, Fifth Edition

Figure 7.1

Function with period 2n.

P

If f (x) has period p, then the substitution: t x=pr*. Yur

~ T T

3-e-

&/

- .

(7.4)

converts f (x) into a function of t having period 2n; for when t increases by 2n, x increases by p. A function f (x) having period 2n is illustrated in Fig. 7.1. Such periodic functions appear in a great variety of physical problems: the vibrations of a spring; the motion of the planets about the sun; the rotation of the earth about its axis; the motion of a pendulum; the tides and wave motion in general; vibrations of a violin string, of an air column (for example, in a flute); and musical sounds in general. The modern theory of light is based on "wave mechanics," with periodic vibrations a characteristic feature; the spectrum of a molecule is simply a picture of the different vibrations taking place simultaneously within it. Electric circuits involve many periodically varying variables, for example, the alternating current. The fact that a journey around the globe involves a total change in longitude of 360" is an expression of the fact that the rectangular coordinates of position on the globe are periodic functions of longitude, with period 360"; many other examples of such periodic functions of angular coordinates can be given. Now it can be shown that every periodic function of x satisfying certain very general conditions can be represented in the form (7.1), that is, as a trigonometric series (see Section 7.3). This mathematical theorem is a reflection of a physical experience most vividly illustrated in the case of sound, for example, that of a violin string. The term ;ao represents the neutral position, the terms a , cos x b1sinx the fundamental tone, the terms a2 cos 2x b2 sin 2x the first overtone (octave); the other terms represent higher overtones. The variable x must here be thought of as time and the function f (x) as the displacement of an instrument, such as a phonograph needle, which is recording the sound, or of a point on the string. Thus the musical tone heard is a combination of simple harmonic vibrations-the terms (a, cos nx b, sin nx). Each such pair can be written in the form

+

+

+

where

A,=,/=,

an=Ansincr,

bn=Ancoso.

The "amplitude" A,+% is a measure of the importance of the nth overtone in the whole sound. The differences in the tones of different musical instruments can be ascribed mainly to the differences in the weights A, of the overtones.

Chapter 7 Fourier Series and Orthogonal Functions

Let us suppose now that a periodic function f ( x ) is the sum of a trigonometric series (7.l), that is, that

What is the relation between the coefficients an and bn and the function f (x)? To answer this, we multiply f (x) by cos mx and integrate from -17 to 17:

If term-by-term integration of the series is allowed, then we find

f ( x ) cos mx dx = (7.6)

The integrals on the right-hand side are easily evaluated with the help of the identities 1

+ y ) + cos (x - y)], 1 sinx cos y = -[sin (x + y ) + sin (x - y ) ] , 2

cos x cos Y = 2 [cos (x

1

sinxsiny = --[cos(x 2

+ y ) -cos(x

(7.7)

- y)].

They give

XomCi

S_:

cos nx cos mx dx =

I

0,

n#m,

n, n = m # O ,

If m = 0, then all terms on the right-hand side of (7.6) are 0 except the first one, and one finds

For any positive integer rn, only the term in a, gives a result different frok 0. Thus

469

470

Advanced Calculus, Fifth Edition Multiplying f (x) by sin mx and proceeding in the same way, we find

From the last three formulas, one now concludes that an=-!

f (x) cos nx dx

(n = 0, 1 , 2 , . . .),

f (x) sin nx dx

(n = 1,2, . . .).

-?r

This is the fundamental rule for coefficients in a Fourier series. Without concerning ourselves with the validity of the steps leading to (7.9) we de$ne a Fourier series to be any trigonometric series

in which the coefficients a,, b, are computed from a function f (x) by (7.9); the series is then called the Fourier series of f (x). Concerning f (x) we assume only that the integrals in (7.9) exist; for this it is sufficient that f (x) be continuous except for a finite number of jumps between -n and n. No parentheses are used in the general definition (7.10). It is common practice to group the terms as in (7.5). However, the series will always be understood in the ungrouped form (7.10). We recall that insertion of parentheses in a convergent series is always permissible (Theorem 27 of Section 6.10).

THEOREM 1 Every uniformly convergent trigonometric series is a Fourier series. More precisely, if the series (7.10) converges uniformly for all x to f (x), then f (x) is continuous for all x , f (x) has period 2n,and the series (7.10) is the Fourier series of f (x). Proof. Since the series converges uniformly for all x, its sum f (x) is continuous for all x (Theorem 3 1 of Section 6.14). The series remains uniformly convergent if all terms are multiplied by cos mx or by sin mx (Theorem 34 of Section 6.14). Accordingly, the term-by-term integration of Eq. (7.6) is justified (Theorem 32 of Section 6.14); (7.9) now follows as previously so that the series is the Fourier series of f (x). The periodicity of f (x) is a consequence of the periodicity of the terms of the series, as remarked in Section 7.1.

COROLLARY If two trigonometric series converge uniformly for all x and have the same sum for all x: 1

-a0

2

00

b0

n=l

n=l

1 + x ( n ncos nx + b, sin nx) -- -ah + x ( a , cos nx + bk sinnx), 2

then the series are identical: a. = ah, a, = a;, b, = b; for n = 1,2, . . . . In particular, if a trigonometric series converges uniformly to 0 for all x, then all coefficients are 0.

471

Chapter 7 Fourier Series and Orthogonal Functions

ProoJ: Let f (x) denote the sum of both series. Then by Theorem 1, &

f(x)cosnxdx

( n = O , l , 2 ,...),

and similarly b, = bl, for all n. If f (x) r 0, then all coefficients are 0.

Although the Fourier series of f (x) is well defined when f (x) is merely "piecewise continuous," it is too much to expect that the series will converge to f (x) under such general conditions. However, it turns out that very little more is required to ensure convergence to f (x). In particular, if f is periodic with period 2n and has continuous first and second derivatives for all x, then the Fourier series of f ( x ) will converge uniformly to f (x) for all x. The result is in itself remarkable when one considers the fact that expansion of f in a convergent power series requires continuous derivatives of all orders-plus the condition that the remainder R, of Taylor's formula converg'es to 0. One can even go further and guarantee uniform convergence of the Fourier series of f (x) to f (x) when f (x) has "corners," that is, points at which fl(x) has a jump discontinuity, while f has continuous first and second derivatives between the corners; this is illustrated in Fig. 7.2. Indeed, one can enlarge the concept of comer to include jump discontinuities off (x), as illustrated in Fig. 7.3; one can hardly expect convergence of the series to f (x) at the discontinuity points, where f (x) may even be ambiguously defined. But the Fourier series makes up our minds for us, in a most reasonable way: It converges to the average of the left- and right-hand limits, that is, to the number

A[2

Figure 7.2

lim f ( x ) + lim f(x)]

x+x1-

x-+x1+

Piecewise very smooth continuous function.

I

Figure 7.3

Piecewise very smooth function with jump discontinuities.

472

Advanced Calculus, Fifth Edition

Figure 7.4

Periodic extension of a function defined between -n and n;

7

*

-

at the discontinuity X I . One cannot expect the series to converge uniformly near the discontinuity (cf. Section 6.14), but it will converge uniformly in each closed interval containing no discontinuity. Although we have up to this point considered only periodic functions f ( x )(with period 2 n ) , it must be remarked that the basic coefficient formulas (7.9)use only the values o f f ( x ) between -IT and JT . Thus if f ( x ) is given only in this interval and is, for example, continuous, then the corresponding Fourier series can be formed, and we continue to call the series the Fourier series of f ( x ) . If the series converges to f ( x )between -n and n , then it will converge outside this interval to a function F(x), which is the "periodic extension of f (x)"; this is illustrated in Fig. 7.4. It should be noted that unless f ( n ) = f ( - n ) , the process of extension will introduce jump discontinuities at x = n and x = - n . At these points the series will converge to the number midway between the two "values" of F ( x ) . We term a function f ( x ) , defined for a 5 x 5 b, piecewise continuous in this interval if the interval can be subdivided into a finite number of subintervals, inside each of which f ( x ) is continuous and has finite limits at the left and right ends of the interval. Accordingly, inside the ith subinterval the function f ( x ) coincides with a function fi(x) that is continuous in the closed subinterval; if, in addition, the functions fi(x) have continuous first derivatives, we term f ( x ) piecewise smooth; if, in addition, the functions f,( x )have continuous second derivatives, we term f ( x ) piecewise very smooth.

FUNDAMENTALTHEOREM Let f ( x )be piecewise very smooth in the interval - n ( x 5 n . Then the Fourier series of f ( x ) : W

2

on =

1J

+ Z(U,, cos nx + bn sinnx), n=l

R

f ( x ) cos nx d x ,

-R

b,, =

Jr

f ( x ) sinnx d x .

-R

converges to f ( x ) wherever f ( x ) is continuous inside the interval. The series converges to

'2[

lim f ( x ) + lim f ( x ) ] x+x1+

x-bx,-

at each point of discontinuity xl inside the interval, and to

1 -[ 2 lim f ( x ) + X-b-R+ lim f ( x ) ] x-+~-

Chapter 7 Fourier Series and Orthogonal Functions at x = fn. The convergence is uniform in each closed interval containing no discontinuity. This theorem is adequate for most applications: The hypotheses can be weakened, "very smooth" being replaced by "smooth; in fact, f (x) need only be expressible as the difference of two functions, both of which are steadily increasing as x increases. For extensions to these cases the reader is referred to the books of Jackson and Zygmund listed at the end of the chapter. The proof of the fundamental theorem will be given in Section 7.9. '-,

We now proceed to consider several examples that will bring out more clearly the relation between f ( x ) and its Fourier series.

EXAMPLE 1 Let f ( x ) have the value -1 for -n Ix < 0 and the value +1 for 0 5 x 5 n. The periodic extension of f ( x ) then gives the "square wave" of Fig. 7.5. One finds I

Figure 7.5

Representation of square wave by Fourier series.

473

474

Advanced Calculus, Fifth Edition Hence for 0 < Ix 1 < n ; : ~ i! 0

,.

4 4 . 4 f(x)= - s i n x + - s i n 3 x + . . . = - ~ n 3n

" sin (2n - 1)x 2n-1

n=l

'

The figure shows the first three partial sums

4 S1=-sinx, n

4 S2=S1+-sin3x,

4 . S3=S2+-sink 5n

3n

of this series. If the graphs are studied carefully, then it becomes clear that f (x) is being approached as limit. For x = 0, each partial sum equals 0, so that the series does converge to the average value at the jump; there is a similar situation at x = fn . However, it should be noted that the approximation to f (x) by each partial sum is poorest immediately to the left and right of the jump points.

+

EXAMPLE 2 Let f (x) = i n x for -n 5 x 5 0 and f (x) = i n - x for 0 ix 5 n. The periodic extension of f (x) is the "triangular wave" of Fig. 7.6. In this example the extended function is continuous for all x. One finds

IS0 -, (;+x)cosnxdx+-

a,, = n

=

n

[ i ( l - cosnn) n2

1 + -(I n2

1"(5 o

- X)

]

cos nx cix

2 n2n

- cosnn) = -(I

- cosnn)

for n = 1 , 2 , . . . . For n = 0 a separate computation is needed:

The computation of the b,'s is like that of the an's, and one finds b, = 0 for n = l , 2 , .... Hence

-C" cos(2n(2n - l)x

4 4 4 f(x)= - c o s x + - c o s 3 x + ~ ~ ~ = n 9n IT

n=1

-

The first two partial sums are plotted in Fig. 7.6. Since there are no jumps, one must expect convergence everywhere. It should, however, be noted that at the corners (where f '(x) has a jump), the convergence is poorer than elsewhere. Thus far we have proceeded formally, evaluating coefficients and verifying graphically that the series converges to the function. We now proceed to examine the steps more carefully. The constant term ao/2 of the series is given by the formula

Chapter 7 Fourier Series and Orthogonal Functions

Figure 7.6

475

Representation of triangular wave by Fourier series. -

The right-hand member is simply the average or arithmetic mean of f (x) over the interval -n 5 x 5 n (Section 4.1). One can also write

In words: The line y = ;ao must be such that the area between the line and the curve y = f (x) lying above the line equals the area between the line and the curve y = f (x) lying below the line. Thus the line y = ;ao is a sort of symmetry line for the graph of y = f (x). From either of these points of view it is clear that in the two examples considered, one must have ;a0 = 0; the average of f (x) is 0, and there is as much area above the x-axis as below. Still another point of view yields the same formula for :ao. We define the total square error of a function g(x) relative to f (x) as the integral

This error is 0 when g = f (or when g = f except for a finite number of points), is otherwise positive. We now seek a constant function y = go such that this error is as small as possible. In other words, we try to approximate y = f (x) as accurately

-

476

Advanced Calculus, Fifth Edition as possible, in terms of least square error, by a constant go. The error is now

where A and B are constants. Thus E(go) is a quadratic function of go, having a minimum when dE/dgo = 0:

Hence the error is minimized when

Thus the constant function y = l a is the best constant approximation, in the sense 2: O of least square error, to the function f (x). This last point of view holds for the coefficients of the general partial sum:

THEOREM 2 Let f (x) be piecewise continuous for -x 5 x 5 n.The coefficients . of the partial sum

of the Fourier series off (x) are precisely those among all coefficients of the function gn(x) = p o + p l c o s x + q l sinx + . - - + p n c o s n x +qnsinnx that render the square error

a minimum. Furthermore, the minimum square error En satisfies the equation: E. = [ : [ f ( x ) l 2 d x - n ."r?r)?

2

+2 (a: + b:) k=l

I

(7.12)

.

74..

The proof is left to Problem 7.

COROLLARY If f (x) is piecewise continuous for -n 5 x 5 n and ao, a l , . . . , bl, bz, . . . are the Fourier coefficients of f (x), then

so that the series C E l ( a i

+ b:)

converges. Furthermore,

lim an = 0,

n-boo

lim bn = 0.

n-b w

,

.

--

477

Chapter 7 Fourier Series and Orthogonal Functions

I Figure 7.7

L

Graphically determined error.

Proofi Since the square error [( f - g)2 d x is always positive or 0 , the minimum square error En is always positive or 0 . Accordingly, (7.13)follows from (7.12).By Theorem 7 of Section 6.5 the series C ( a i b i ) must converge or diverge properly; because of (7.13), the series cannot be properly divergent. Therefore the series converges; (7.14)then follows from the fact that the nth term of the series converges to 0.

+

It is shown in Section 7.12 that En can be made as small as desired by choosing n sufficiently large; that is, the sequence En converges to 0 . The relation (7.13) is Bessel 's inequality. Theorem 2 can be made the basis of a graphical estimation of Fourier coefficients. Thus for the function of Example 1, one first chooses the best-fitting constant term ;ao; this is clearly 0. One then tries to add a function pl cos x + p 2 sin x to make the square error as small as possible. It is apparent that the best approximation is achieved by t a l n g a sine term alone. The function sin x itself fits fairly well, though the errors are large near fn and 0 . To reduce these, we overshoot at in,taking, say, 1.3 sinx. New errors are introduced near i n , but the total square error will be less. If we subtract 1.3 sinx from f (x) graphically, we find the function of Fig. 7.7. To eliminate this error, it is clear that a function p6 sin 3x is called for. Again we overshoot and estimate p6 = 0.4. We subtract 0.4 sin 3x from the function graphed in Fig. 7.7 and so on. We thus obtain the expression 8. f ( x ) = 1.3 sin x

+ 0.4 sin 3x.

..'

t

.s

This agrees, up to the number of significant figures carried, with the first two terms of the Fourier series

4

- sinx + L s i n 3 x n 3n

+ . a .

obtained above. It is recommended that the following problems be solved first roughly by this graphical procedure. In this way, considerable insight into the structure of the series can be gained. In constructing the series in this way it is best to think of the pair a, cos nx b, sin nx in the amplitude-phase form:

+

a, cos nx

+ b, sin nx = A, sin (nx + a)

(7.15)

Advanced Calculus, Fifth Edition pointed out at the end of Section 7.1. The phase angle cr effectively determines how much the curve y = sin n x has to be shifted to the left or right to match the given oscillation; the amplitude A, merely adjusts the vertical scale. It should be remarked that the process of decomposing a periodic phenomenon into its component simple harmonic parts is used in a great variety of common experiences. The rattling of an old-fashioned vehicle corresponds to a high-frequency (large n ) component with large amplitude; we automatically separate this from a lowfrequency vibration, or swaying. In a less precise sense, a large day-to-day fluctuation in weather conditions also corresponds to a high-frequency component of large amplitude; the seasonal changes are of low frequency and are much less disturbing.

PROBLEMS 1. Find the Fourier series for each of the following functions:

+$-g,

I) G ( X ) = ; - 5 - $ , IT - x 5 ~ 5 0 ; G ( x ) = f 0sxsn (It is suggested that the first few partial sums be graphed and compared with the function in each case.) 2. It follows from the fundamental theorem of Section 7.3 that if f (x) is defined between 0 and 2n and is piecewise very smooth in that interval, then f (x) can be represented by a series of form (7.1) in that interval. a) Show that the coefficients a, and b, are given by the formulas: 2n 1 2n b,,=--1 f (x) sin nx dx. a,=L/ f(x)cosnxdx. x 0 b) Extend this result to a function defined from x = c to x = c + 2n, where c is any constant. 3. Using the results of Problem 2(a), find Fourier series for the following functions: 4. Determine which of the following functions are periodic and find the smallest period of those that are periodic: a) sin5x b) cos 5 c) sinxx . f ) sin $ sin d) x sinx e) sin 3x + sin 5x g) sin x + sin xx 5. The result of Problem l(a) implies that for 0 < x < x , sin 3x sinx+ -+... 3

+ 4

Use x = i n in this equation to show that

479

Chapter 7 Fourier Series and Orthogonal Functions 6. Use the method of Problem 5 to obtain, with the aid of the series developed in Problems 1 and 3, the relations a) R = l + 3 - 15 - 71+ 9 1+ n 1- - - -1 + + 1. . . 1

2/8

...+ ( Z n -I l ) T + ... b) % = 17 +17 +1+ + 2

p 1 + ~ -1$ + " .

,)+I-. d)

c

1 1 $=++i2+F+-jf+...

7. Prove Theorem 2. [Hint: Show that

+ . .. +

nq, - 29, (

2

S

1

f sinnx d x ;

accordingly, if po, pl , . . . , qn are chosen to give each term on the right its smallest value, -, the error will be minimized.] .j 8. a) Prove the trigonometric identities:

b) Obtain analogous expressions for sinnx and cosn x . [Hint: Use the identities sinx = :(eix - e-'~)/i,cos x = i(eix + e-iX).] C) Show that the identities of parts (a) and (b) can be interpreted as Fourier series expansions.

The Fourier series up to this point have been considered only for functions of period 2n or, more restrictedly, for functions defined between -n and n.We now proceed to enlarge the scope of the theory. If f ( x ) is a function of period 2n,one can use as basic interval any interval c _( x 5 c 2n,that is, any interval of length 2n.For such an interval the same reasoning as previously leads to a Fourier series

+

(10 2

+ p ( a n cos n x + bn sin n x ) , n=l

where

1

c+2n

f ( x ) sin nx d x . bn=--l If f ( x ) is given for all x , with period 2n,this is merely another way of computing the coefficients a,, b,. If f (x) is given only for c 5 x 5 c 2n,the series can be used to represent f in this interval; it will then (if convergent) represent the periodic extension of f outside this interval.

+

Advanced Calculus, Fifth Edition The interval -n 5 x 5 n has certain advantages for utilization of symmetry properties. Let f (x) be defined in this interval and let f (-x) = f (x),

- n 5 x 5 IT.

(7.17)

Then f is called an even function of x (in the given interval). If, on the other hand, then f is called an odd function of x. We note that the product of two even functions or of two odd functions is even whereas the product of an odd function and an even function is odd. Furthermore, f odd

21: f(x)dx,

feven.

Let f now be even in the interval -n 5 x 5 n. Then f (x) cos nx is even (product of two even functions) whereas f (x) sin nx is odd (product of odd function and even function). Hence by (7.19), an =

?/" 15

b,=O

f(x)cosnxdx o ( n = 1 , 2 ,...).

Similarly, i f f is odd, 2 b,, = -

a,, = 0,

1"

(n = 0 . 1.2. ...),

f (x) sin nx d x .

(7.20)

(7.21)

0

We have thus the expansions (for a function piecewise very smooth):

f (x) =

CC

$+

ancos nx

(f even),

n=l

a,, =

2n J" f(x) cos nx dx 0

and w

f (x) =

b, sin nx n=l

b,, =

2 IT

/"

(f odd),

(x) sinnx dx.

0

Now (7.22) uses only the values of f (x) between x = 0 and x = n. Hence for any function f (x) given only over this interval, one can form the series (7.22). This is called the Fourier cosine series of f (x). It follows from the fundamental theorem that the series will converge to f (x) for 0 5 x 5 n and outside this interval to the even periodic function that coincides with f (x) for 0 5 x 5 n.This is illustrated in Fig. 7.8. In the same way, (7.23) defines the Fourier sine series of a function f (x) defined only between 0 and n. The series represents an odd periodic function that coincides with f (x) for 0 < x < n,as illustrated in Fig. 7.9.

Chapter 7 Fourier Series and Orthogonal Functions

Figure 7.8

Even periodic extension of function defined between 0 and n.

Figure 7.9

Odd periodic extension of function defined between 0 and n .

EXAMPLE Let f ( x ) = n - x . Then one can represent f ( x ) by a Fourier series over the interval -n c x < n . The formulas (7.9) give

Hence one has 00

n - x = n + z Y

(- 1)" sin nx

, -n<x
The same function, n - x , can be represented by a Fourier cosine series oveii' the interval 0 5 x Irr. The formulas (7.22) give

Hence one has

=

-+2

n 2(

2cosx+-

2 cos 3x 2 cos 5x +-+... 32 52

). O j x s n .

(b)

481

482

Advanced Calculus, Fifth Edition

Figure 7.10 series.

(a) Fourier series. (b) Fourier cosine series. (c) Fourier sine

,

FinaIly, the same function, n - x , can be represented by a Fourier sine series over the interval 0 < x t n . The formulas (7.23) give '

,U =

00

r r - x = 2 y - sin nx

.

Ocxcn.

Figure 7.10 shows the graphs of the three functions represented by the series (a), (b)?(c). Change of period. If f (x) has period p:

then the substitution

transforms f ( x ) into a function g(t):

and g(t) has period 217, since

,.

,.

483

Chapter 7 Fourier Series and Orthogonal Functions The change from x to t is simply a change of scale. Since g has period 2 n , one has a Fourier series for g (assumed piecewise very smooth): 00

g(t) = 5 2

+ x ( a ncos nt + bn sin n t ) , n=l ?*t&&

where, for example, an =

/"

g ( t )cos nt d t ,

3C.

-H

1 bn = -

n

/"

g(t) sin nt d t .

-H

.-!

If now t is replaced by (2n/p)x, one finds a Fourier series for f (x):

f (x) =

7 +2

0 s (n .

n=l

Fx)+

bn sin ( n .

Fx)] .

(7.24)

The coefficients a n ,bn can be expressed directly in terms of f (x). Thus:

where p = 2L. The Fourier cosine series can also be used in this case. One finds

where

Similarly, f (x) has a Fourier sine series:

where

+

EXAMPLE Let f (x) = 2x 1. Then f (x) can be represented by a Fourier series ' over the interval 0 c x c 2. Here p = 2 and t = n x , so that

484

Advanced Calculus, Fifth Edition where g(t) is defined for 0 < t 5 277. Hence g(t) can be represented by a Fourier series: 00

g(t) =

2

+ x ( a n cos nt + b, sin nt), n=l

where

r-.

a , = L n1 2 n ( $ + I ) c o s n t d t ,

bn=;

' lfr :( +

1) sin nt d t .

One finds a0=6,

al = a 2 = ... = 0,

b,=--

4 nn'

s

so that

and hence

.. One could use the formulas (7.25) directly, but the change to t simplifies the calculations.

PROBLEMS

+

1. Let f ( x ) = 2x 1 . a) Expand f ( x ) in a Fourier series for -n < x < n. b) Expand f ( x ) in a Fourier series for 0 < x < 2 n . c) Expand f ( x ) in a Fourier cosine series for 0 5 x 5 n . d) Expand f ( x ) in a Fourier sine series for 0 < x < n. e) Expand f ( x ) in a Fourier series for 0 < x < n. f ) Graph the functions represented b y the series of parts (a), (b), (c), (d), and (e). 2. Let f ( x ) = x2. a) Expand f ( x ) in a Fourier series for 77 < x < 3 n . b) Expand f ( x ) in a Fourier series for 1 < x < 2. c ) Graph the functions represented b y the series of parts (a) and (b). 3. Let f ( x ) = sin x . a) Expand f ( x ) in a Fourier series for 0 5 x _( 2 n . b) Expand f ( x ) in a Fourier series for 0 5 x 5 n . c) Expand f ( x ) in a Fourier cosine series for 0 5 x 5 n.

Chapter 7 Fourier Series and Orthogonal Functions 4. Let f ( x ) = x . a) Expand f ( x ) in a Fourier cosine series for 0 5 x 5 n . b) Expand f ( x ) in a Fourier sine series for 0 5 x < n. C ) Expand f ( x ) in a Fourier cosine series for 0 5 x 5 1 . d) Expand f ( x ) in a Fourier sine series for 0 5 x < 1 . 5. Let f ( x ) = C z ,(a, cos n x b, sin n x ) , where the series converges uniformly for all x . State what conclusions can be drawn concerning the coefficients a,, b, from each of the following properties of f ( x ) :

9+

= f (x) C) f ( n -x ) = f (x) e) f (-XI = f ( x ) = f g) f ( n X ) = f (XI i) +x) =f(x) a)

+

f (-x)

+ f(5

f (-XI = -f ( x ) d) f (5 - x ) = f ( x ) f ) f (71.- X ) = - f (XI h) f (%+ x ) = f ( x ) j) f ( x ) = f (2x1 b)

(5 - x )

[Hint: Use the corollary to Theorem 1.]

As was indicated in Section 7.1, the natural field of application of Fourier series is to periodic phenomena. The fact that a periodic function f ( t ) can be resolved into its simple harmonic components A, sin (nt a,) is of fundamental physical significance. For all "linear" problems this resolution permits one to reduce the problem to the simpler one of a single simple harmonic vibration and then to build up the general case by addition (superposition) of the simple ones. The concrete application of Fourier series to such problems takes two main forms: A periodic function f (t) may be given in graphical or tabulated form; an understanding of the physical mechanism leading to such a function requires a "harmonic analysis" of f (t), or representation of f (t) as a Fourier series. Second, the function f ( t ) is known to be periodic and is known to satisfy some implicit relation, such as a differential equation; it is desired to determine f (t) as a Fourier series on the basis of this information. The first problem is thus one of interpretation of experimental data; the second is one of prediction of the result of an experiment, on the basis of a mathematical theory. Although the application of Fourier series to periodic phenomena is basic, there is a much wider field of application. As has been shown, an "arbitrary" function f ( x ) , given for a 5 x 5 b, has a representation as a Fourier series over that interval. Thus in any problem concerning a function over an interval it may be advantageous to represent the function by the corresponding series. This permits a tremendous variety of applications. As before, the applications usually take the form of interpretinggiven data or of predicting functions satisfying given conditions. In this book, particular applications will be considered in Chapter 10 on partial differential equations. In the following problems, several illustrations of the form of solutions to differential equations are provided.

+

485

486

Advanced Calculus, Fifth Edition

,$I.T:

PROBLEMS 1. Show that the linear differential equation

is satisfied by -

>=

y = Acoswt

+ Bsinot,

where

and determine a solution, in series form, of the differential equation

d2y

00

dy + y + 44b

+

= f (t) = x ( a n cosnt bn sinnt). n=l 2. Show that, granting the correctness of the necessary term-by-term differentiations of series, the function CO

+

f(x, t ) = C [ A , cosncr B, sinnct] sinnx. n=l where the A, and Bn are constants, satisfies the partial differential equation

a2f -

2

R

a2f

= -,a x 2 The differential equation is that of the vibrating string, and the series represents the general solution when the ends are fixed, rr units apart. 3. Show that, granting the correctness of the necessary term-by-term differentiations of series, the function at2

00

f (r, 8 ) = A0

+ z ( ~ , r "cos no + B.r"

sin no)

-

n=l satisfies the Laplace equation in polar coordinates:

As was shown in Section 5.15, this equation describes equilibrium temperature distributions, electrostatic potentials, and velocity potentials. Every function f that is harmonic in a circular domain r c R can be represented by such a series, as will be shown in Section 9.1 1.

If two functions f ( x ) and f i ( x ) have the same set of Fourier coefficients:

are the functions necessarily identical? In other words, is a function uniquely determined by its Fourier coefficients? The answer is in the affirmative:

487

Chapter 7 Fourier Series and Orthogonal Functions

Figure 7.11

Pulse function.

i, THEOREM 3 (Uniqueness Theorem) Let f ( x ) and f , ( x ) be piecewise continuous in the interval -n 5 x 5 n and satisfy (7.30),so that the two functions have the same Fourier coefficients. Then f ( x ) = f l ( x ) except perhaps at points of discontinuity. Proof. Let h ( x ) = f ( x ) - f i ( x ) . Then h ( x ) is piecewise continuous, and from (7.30)it at once follows that all Fourier coefficients of h ( x ) are 0. We then show that h ( x ) = 0 except perhaps at discontinuity points. Let us suppose h(xo) # 0 at a point of continuity xo, for example, h(xo)= 2c > 0 . Then, by continuity, h ( x ) > c for ( x - xol < 6 and 8 sufficiently small (cf. Problem 7 following Section 2.4). We can assume -n < xo < n . We now achieve a contradiction by showing that there exists a "trigonometric polynomial"

.

P ( x ) = po

+ p1 cosx + pq sinx + . . - + p

~ cos - kx~

<

+ p ~ sin kx

that represents a "pulse" at xo of arbitrarily large amplitude and arbitrarily small width. This is pictured in Fig. 7.1 1 . If such a pulse can be constructed, then one has a contradiction. On one hand, t

h ( x ) P ( x ) d x = po

h(x)dx

+ pt

+.

~ ( X ) C O dSxX

=0.

On the other hand, the major portion of the integral j h ( x ) P ( x )dx is concentrated in the interval in which the pulse occurs; here h ( x ) is positive, and P ( x ) is large and positive. Hence the integral is positive and cannot be 0. ~i To make this precise, we take

P(x) = [+(x)lN,

+(x) = 1

+ cos ( X - x0) - cos 8

' 2

03

488

Advanced Calculus, Fifth Edition for an appropriate positive integer N. Since the functions sinnx and cosnx are expressible as trigonometric polynomials (Problem 8 following Section 7.4), the function P(x) is a trigonometric polynomial. Let

Then k > 1 and P 2 kN for (x - xol 5 i6. Since $ is positive (greater than 1) for IS 2 5 Ix - xoI < 6, P is positive in this range. On the other hand, I $(x)l < 1 for -n 4 x < x o - 6 a n d f o r x o + 6 < x ( n , s o t h a t IPI < 1 inthis range. Now the function h(x), being piecewise continuous, is bounded by a constant M for -n 5 x 1 n:Ih(x)I IM. It follows from the properties of P(x) just listed that P(x)h(x) > - M for -n I x 5 xo - $5 and forxo $3 (x 5 n , while P(x)h(x) 2 ckN for xo - i S (x _( xo $5. Accordingly, by rule (4.17) of Section 4.1,

+

+

Since kN -+ +oo as N -+ oo, the right-hand member of the inequality is surely positive when N is sufficiently large. Accordingly, the left-hand member is positive for appropriate choice of N. This contradicts the fact that the left-hand member is 0. Accordingly, h(x) = f (x) - f i (x) = 0 wherever f (x) and f i (x) are continuous. Remarks. The uniqueness theorem can be looked at in another way, namely, as asserting that the system of functions

1,

cos x,

sin x , . . . , cos nx,

sin nx, . . .

is "large enough," that is, that there are enough functions in this system to construct series for all the periodic functions envisaged. It should be noted that omission of any one function of the system would destroy this property. Thus if cos x were omitted, one could still form a series 1 -a0 + b l sinx +a2cos2x +b2sin2x + - . . 2 as before. But there are very smooth periodic functions whose Fourier series in this deficient form could never converge to the function, namely, ail functions A cos x for A = const. # 0. For each such function would have all coefficients 0:

S_:

~ c o s x d x= 0,

J_: A cosx sinx dx = 0,

. .. .

The series reduces to 0 and cannot represent the function. The essence of the proof of the uniqueness theorem is the demonstration that there are enough functions in the system of sines and cosines to construct a pulse function P(x).

THEOREM 4 Let the function f (x) be continuous for -n 5 x 5 n and let the Fourier series of f (x) converge uniformly in this interval. Then the series converges to f(x) for -n ( x s n .

'

Chapter 7 Fourier Series and Orthogonal Functions

Proof. Let the"3umof the Fourier series of f ( x ) be denoted by f i ( x ) :

Since the series converges uniformly, it follows from Theorem 1 that f i ( x ) is continuous and that a , , bn are the Fourier coefficients of f i ( x ) .But the series was given as the Fourier series of f ( x ) . Hence f ( x ) and f ,( x ) have the same Fourier coefficients, and by Theorem 3, f ( x ) = f l ( x ) ; that is, f ( x ) is the sum of its Fourier series for -n F x _ < n .

THEOREM FOR CONTINUOUS, 7.8 PROOFOF FUNDAMENTAL PERIODIC, AND PIECEWISE VERYSMOOTH FUNCTIONS

THEOREM 5 Let f ( x )be continuous and piecewise very smooth for all x and let f ( x ) have period 2 n . Then the Fourier series of f ( x ) converges uniformly to f ( x ) for all x .

Proof. Let us first assume that f ( x ) has continuous first and second derivatives for all x . One has (for n # 0) an =

A n

/"

f ( x ) cosnx d x =

f ( x ) sin nx nn ,

-R

1:. ~1~ 1

-

"

f ' ( x )sin nx d x

by integration by parts. The first term on the right is zero. A second integration by parts gives a, =

,:I

f ' ( x )cos nx n2n

1

/_" f " ( x )cos n x d x

- ;;i;;

R

1"

f " ( x )cos nx d x , = -n2n - ,

the first term being zero because of the periodicity of f ' ( x ) .The function f " ( x ) is continuous in the interval -n 5 x 5 n,and hence If "(x)l ( M for an appropriate constant M . One concludes that

n all ~ n . Hence each term of In exactly the same way we prove that (bn(5 2 ~ / for the Fourier series of f ( x ) is in absolute value at most equal to the corresponding term of the convergent series

Application of the Weierstrass M-test (Section 6.13) now establishes that the Fourier series converges uniformly for all x . By Theorem 4 the sum is f ( x ) . We now consider the case of a function f ( x ) that is periodic, continuous, and piecewise very smooth. The only step that requires reexamination is the proof that la, I ( 2 ~ n and - ~that Ib, I 5 2 M n w 2 .The integration by parts must now be carried

489

490

Advanced Calculus, Fifth Edition out separately ovk? each interval within which f "(xJis continuous. If.tHereBult3 are" added, one obtains, for example, for b,:

The integrals are technically improper but exist as such. Since f is continuous and periodic, the terms in the first bracket add up to -f ( x ) cos nx

The functions f ' ( x )and f " ( x )are bounded in each subinterval. Hence one constant M can be chosen so that If ' ( x ) J5 M and I f "(x)l IM throughout. If there are k subintervals, we conclude that

A similar result holds for a,. Hence the Fourier series of a periodic, continuous, piecewise very smooth function converges uniformly to the function for all x.

Before proceeding to the general case of a function with jump discontinuities we consider an example illustrating the result just obtained. Let

and let G be repeated periodically outside this interval, as shown in Fig. 7.12. The resulting function G ( x ) is continuous for all x and is piecewise smooth. Its

I

I

,

I

0

r-

Figure 7.12

Auxiliary function Wx).

I I

K

8

-x

Chapter 7 Fourier Series and Orthogonal Functions Fourier series is found in Problem l(f) following Section 7.4 to be the series

2n 1 " cosnx ---C-. 3

I7 n=1

n2

Hence lan[ 5 ~ n as- asserted, ~ with M = 1/17; the b, happens to be 0. By Theorem 5 this series converges uniformly to G(x). We now ask: Is term-by-term differentiation of the series permissible? In other words, is

wherever G1(x) is defined? By Theorem 33 of Section 6.14 this is correct if x lies within an interval within which the differentiated series converges uniformly. It is shown in Problem 4 below that the series C(sinnx/n) converges uniformly for a 5 1x1 (n, provided that a > 0. Accordingly, (7.31) is correct for -n 5 x 5 17, except for x = 0. Now let F(x) be the periodic function of period 217 such that F(0) = 0 and

The function F(x) is shown in Fig. 7.13. We have now proved that

I

Figure 7.13

Removal of jump discontinuity.

491

492

Advanced Calculus, Fifth Edition

!-rut

for all x , the convergence being uniform for 0 < a 5 Ix I 5 n. The series on the righthand side was computed as the Fourier series of F ( x ) in Problem l ( e ) following Section 7.4. Accordingly, we have shown that F ( x ) is represented by its Fourier series for all x . The remarkable feature of this result is that F ( x ) has a jump, from - to at x = 0. The series converges to the average value F(0) = 0. We have therefore verified a special case of the following theorem:

&

THEOREM 6 Let f ( x ) be defined and piecewise very smooth for -n 5 x 5 n and let f ( x )be defined outside this interval in such a manner that f ( x )has period 2n. Then the Fourier series of f ( x )converges uniformly to f ( x )in each closed interval containing no discontinuity of f ( x ) . At each discontinuity xo the series converges to i T

21

[

$1 "

lim f ( x ) 2 x-+xo+

+ x-bxolim

f (x)].

Prooj For convenience we redefine f ( x )at each discontinuity xo as the average of left and right limit values. Let us suppose, for example, that the only discontinuity is at x = 0 (and the points 2kn, k = f1, f2, . . .) as in Fig. 7.13. Let lim f (x) - lim f ( x ) = S ,

x+o+

x+o-

so that s is precisely the "jump." We then proceed to eliminate the discontinuity by subtracting from f ( x )the function s F(x),where F ( x ) is the function defined earlier. Since s F ( x ) has also the jump s at x = 0 (and x = 2kn), g ( x ) = f ( x )- s F ( x ) has jump 0 at x = 0 and is continuous for all x ; for lim g(x) = lirn f ( x )- s lim F(x)

x-0-

x-bo-

x-0-

and a similar statement applies to the right-hand limit. Since F ( x )is piecewise linear, g ( x ) is continuous and piecewise very smooth for all x and has period 2n. Hence by Theorem 5, g ( x ) is representable by a uniformly convergent Fourier series for all x : 1 g ( x ) = -Ao 2

00

+ X(A.cos nx + B,, sin nx). n=l

Accordingly,

+

1 2

00

s sinnx + C(A, cos nx + B, sin n x ) + - C n 00

f ( x ) = g(x) s ~ ( x=) - A ~

n=l

IT n=l

Chapter 7 Fourier Series and Orthogonal Functions so that f ( x ) is represented by a trigonometric series for all x . The series is the Fourier series of f ( x ) , for 1

bn = -

=

1 -3T

f (x) sin nx dx = -

=

S"

[ g ( x )+ s F ( x ) ]sin nx d x

-7r

and similarly, a, = A,. Therefore the Fourier series of f ( x ) converges to f ( x ) for all x . At x = 0 the series converges to f ( 0 ) , which was defined to be the average of left and right limits at x = 0. Since the series for g ( x ) is uniformly convergent for all x , while the series for F ( x ) converges uniformly in each closed interval not containing x = 0 (or x = 2kn), the Fourier series of f ( x ) converges uniformly in each such closed interval. The theorem has now been proved for the case of just one jump discontinuity. If there are several jumps, at points x l , x2, . . . , we simply remove them by subtracting from f ( x ) the function

The resulting function g ( x ) is again continuous and piecewise very smooth, so that the same conclusion holds. Thus the theorem is established in all generality. Remarks. The proof just given uses the principle of superposition: The Fourier series of a linear combination of two functions is the same linear combination of the corresponding two series. This can be put to very good use to systematize the computation of Fourier series. Illustrations are given in Problem 1 which follows. The idea of subtracting off the series corresponding to a jump discontinuity also has a practical significance. If a function f ( x ) is defined by its Fourier series and is not otherwise explicitly known, one can, of course, use the series to tabulate the function. If f ( x ) has a jump discontinuity, the convergence will be poor near the discontinuity; this will reveal itself in the presence of terms having coefficients approaching O like l l n . If the discontinuity xl and jump sl are known, as is often the case, one can subtract the corresponding function sl F ( x - X I ) as before; the new series will converge much more rapidly. The same idea can be applied to functions f ( x ) that are continuous but for which f l ( x ) has a jump discontinuity sl at x l . One now subtracts from f the function s l G ( x - x l ) , for this continuous function has as derivative precisely the function sl F ( x - x l ) , with jump sl at xl . By integrating G ( x ) - 21713, one obtains a periodic continuous function having a jump in its second derivative at x = 0. Continuing in this way, one provides a jump function for each derivative; each such function can be used to remove corresponding slowly converging terms from the Fourier series.

PROBLEMS 1. Let f i ( x ) and f2(x) be defined by the equations: fi(x) = 0, - x 0 f i ( x ) = 1, O i x l n ; f 2 ( ~=) 0 , - n ~ x < O ; f 2 ( x ) = x , O l x < n .

'

493

494

Advanced Calculus, Fifth Edition Then fl ( x ) and f 2 ( x ) can be represented by Fourier series:

n 2 O0 cos(2n - l ) x f2(x) = - - 4 x n=l (2n ,=I

sin n x C(-1)" n

C

,

-n<x
Withoutfurther integration, find the Fourier series for the following functions: a ) f 3 ( x ) = 1, - n < x < O ; f 3 ( x ) = 0 , O I X ~ ~ ; b) f 4 ( x ) = x , - n ~ x ~ Of 4; ( x ) = 0 , O s x s n ; c) f5(x) = 1, -n 5 x < 0 ; f 5 ( x ) = x , 0 5 x 5 n ; d ) f 6 ( x ) = 2, -15 5 X < 0 ; f 6 ( x ) = 0 , 0 5 x 5 n ; e) f7(x) = 2, -n 5 x < 0 ; f 7 ( x ) = 3, 0 5 x 5 n ; f ) f s ( x ) = 1, -n 5 x 5 0 ; f 8 ( x ) = 1 2 x , 0 5 x 5 n ; g) f 9 ( x ) = a + b x , - n c x < O ; f9(x)=c+dx, O s x i n . 2. Let f ( x ) = (a, cos n x b, sin n x ) , -n 5 x 5 n. By squaring this series and integrating (assuming the operations are permitted) show that

+

9 +xEl

+

[This relation, known as Parseval's equation, can be justified for the most genaal function f ( x ) considered before. See Sections 7.1 1 and 7.12.1 3. Use the result of Problem 2 and the Fourier series found in previous problems to establish the formulas:

n6 = F1 + 217 ; + . . .1+ ; ; i ; + ' . . . d) m

4. Prove that the series sin 2x

sin n x + -+... n

n=l

converges uniformly in each interval -n 5 x 5 a , a 5 x 5 n , provided that a > 0. This can be established by the following procedure: a ) Let p,(x) = sin x + . . - sin n x . Prove the identity:

+

~ n ( x= )

cos i x - cos (n 2 sin i x

+ $)x

( x # 0 , x # f2 n , . . .). [Hint: Multiply p,(x) by sin i x and apply the identity (7.7) for sin x sin y to each term of the result.]

Chapter 7 Fourier Series and Orthogonal Functions b) Show that if a > 0 and a 5 1x1 5 n,then Ipn(x)ls1/ sin

la.

Show that the nth partial sum Sn(x)of the series sin nx sin 2x sinx + -+.-.+-+... \, 2 n can be written as follows PI(X) -+...+p2(x) pn-l(x) pn(x) Sn(x)= 1.2 2.3 n(n - 1) n [Hint: Write sin x = pl , sin 2x = p2 - pl, and so on.] sinnx is . uniformly convergent for a 5 (xI 5 n,where a > 0. d) Show that the series Cgl --, [Hint: By (c), C)

,

+-.

+

Hence uniform convergence of the sequences S,* and pn/n implies uniform convergence of Sn(x).The sequence S,* converges uniformly, since it is the nth partial sum of the series P~(x) ' n=l n(n + 1)' which converges uniformly, because of (b), by the M-test. The sequence pn/n converges uniformly to 0 by (b).]

2

If one reviews the theory of expansion of a function in a Fourier series, as presented in the preceding sections of this chapter, it is natural to ask why the trigonometric functions sin nx and cos nx play such a special part and whether these functions can be replaced by other functions. If one is interested only in periodic functions, there is indeed no natural alternative. However, if one is concerned with representation of a function over a given interval, it will be seen that a great variety of other series is available, in particular, series of Legendre polynomials, Bessel functions, Laguerre polynomials, Jacobi polynomials, Hermite polynomials, and general Sturm-Liouville series. Let f (x) be given in an interval a (x ( b; this interval will be fixed throughout the following discussion. Let ~ $ ~ ( x~) $, ~ ( x.).,. , C$n(x),. . . be functions all piecewise continuous in this interval; this sequence is to replace the system of sines and cosines. We then postulate a development,

just as in the case of Fourier series. Our next step, for Fourier series, was to multiply both sides by cos mx or sin mx and integrate from -n to n; when we did this, all terms dropped out except the one in a, or b,, respectively, because of the relations

495

496

Advanced Calculus, Fifth Edition By analogy we multiply both sides of (7.32) by #,(x) and integrate term by term:

In order to achieve a result analogous to that for Fourier series, we must postulate that L'

The series on the right of (7.33) then reduces to one term:

The integral on the right is a certain constant:

The constant B, will be positive unless 4,(x) = 0 (except at a finite number of points); to avoid this trivial case, we assume that no B, is 0. Then

Thus under the simple conditions (7.34) and (7.36) we have a rule for formation of a series just like that for Fourier series and can hope that analogous convergence theorems can also be proved. We now summarize the assumptions in formal definitions. Dejnitions. Tho functions p(x), q(x), which are piecewise continuous for a 5 x 5 b, are orthogonal in this interval if

A system of functions { ~ $ ~ ( x(n) }= 1,2, . . .) is termed an orthogonal system in the interval a 5 x (b if 4, and 4, are orthogonal for each pair of distinct indices m ,n:

and no qjn(x)is identically 0 except at a finite number of points. 1: f7r

EXAMPLE

The trigonometric system in the interval -n 5 x 5 n: 1,

The function

cosx,

sinx, . . . ,

cosnx,

sinnx, . . . .

is the constant 1; 42is the function cos x, . . . .

Chapter 7 Fourier Series and Orthogonal Functions If f (x) is piecewise continuous in the interval a 5 x 5 b and {$,(x)} is an orthogonal system in this interval, then the series

where

is called the Fourier series o f f with respect to the system {&(x)). The numbers e l ,c2, . . . are called the Fourier coefficients of f (x) with respect to the system {4n(x>l. The preceding formulas can be simplified if one assumes that the constant Bn is always 1, that is, that . t i

This can always be achieved by dividing the original 4,(x) by appropriate constants. When the condition B, = 1 is satisfied for all n, the system of functions @n(x) is called normalized. A system that is both normalized and orthogonal is called orthonormal. This is illustrated by the functions: ,

1

cosx sinx Jn, ,

-

7

cosnx fi 9

sinnx

fi

-9

" "

Although the general theory is simpler for'normalized systems, the advantages f& applications are slight, and we shall not use normalization in what follows. The operations with orthogonal systems are strikingly similar to those with vectors. In fact, we can consider the piecewise continuous functions for a 5 x 5 b as a sort of vector space, as in Section 1.18. The sum or difference of two such functions f (x), g(x) is again piecewise continuous, as is the product cf of f (x) by a constant or scalar c . Equation (7.38) suggests a definition of inner product (or scalar product): !:

One can then define a norm (or absolute value):

The zero function 0* is a function that is 0 except at a finite number of points; in general, in this vector theory of functions we consider two functions that differ only at a finite number of points to be the same function. It is now a straightforward exercise to verify that all the axioms for a Euclidean vector space, except the one

497

Advanced Calculus, Fifth Edition concerning dimension, are satisfied. For convenience we restate the axioms here [cf. (1.102) in Section 1.141.

+ +

11. t f 8) h = f IV. (el + c2)f = C l f VI. 1 . f = f ,

1. f + g = g + f , 111. c t f g ) = cf cg, v. ( ~ 1 ~ z = ) f ~ l ( ~ ),2 f VII. 0 f = o*,

+

+

IX. ( f + g , h ) = ( f , h ) + ( g , h ) , XI. (f,f ) 3 0,

+ (g + h), + c2f ,

VIII. ( f ,g ) = ( g , f ), X. ( c f , g ) = c ( f ,g), XII. ( f , f ) = 0 if and only if

f = O*. (7.44)

The proof is left as an exercise (Problem 3 at the end of this section). We define k functions fi , f2, . . . , fk to be linearly independent if the only scalars cl, . . . , c, for which

are the numbers 0 , . . . , 0. In place of Rule XI11 of Section 1.15 we have now the theorem that for every integer n there exist n linearly independent functions. For example, the functions sin x , sin 2x, . . . , sin nx are linearly independent for -n 5 x 5 n. (See Problem 9 following Section 1.15.) The theorem of Section 1.14 is proved as a consequence of the axioms (7.44) alone. Hence this theorem holds for functions.

I

THEOREM 7

Let f ( x )and g ( x ) be piecewise continuous for a _< x 5 b. Then

I(f, g)l 5

I f 11 . llgll.

(7.46)

The equality holds precisely when f and g are linearly dependent. Furthermore,

Ilf

+ gll 5 l l f ll + llgll;

(7.47)

the equal sign holds precisely when f = cg or g = cf , where c is a positive constant or 0. The relation (7.46) is the Schwarz inequality. If we use the definitions (7.42) and (7.43),it can be written as follows:

The relation (7.47) is the ~in'kowskiinequaliry. In explicit form it reads:

Chapter 7 Fourier Series and Orthogonal Functions

The definition of orthogonal functions can be restated in terms of inner products: f ( x ) is orthogonal to g ( x ) if ( f , g ) = 0. An orthogonal system is a system ($,,(x)} (n = 1 , 2 , . . .) of functions all piecewise continuous for a 5 x 5 b and such that (4m,$n)=O,

m #n;

( 4 n 3 4 n ) = l14n112

= Bn > 0 .

(7.50)

The system is orthonormal if (7.50) holds, and ))@,,I) = 1 for n = 1 , 2 , . . . , so that the @,, are "unit vectors." The Fourier series of f ( x ) with respect to the orthogonal system { & ( x ) ) is the series

If the system is orthonormal, the series becomes

.

which is analogous to the expressions

for a vector in terms of base vectors in 3-dimensional space and n-dimensional space. However, since (7.5 1 ) is an infinite series, complications arise through convergence questions. THEOREM 8 Let {+n(x)) be an orthogonal system of continuous functions for the interval a 5 x 5 b. If the series Czlc n 4 , ( x ) converges uniformly to f ( x ) for alxlb,then

so that the series is the Fourier series of f ( x ) with respect to {c$,(x)].If the system is orthonormal, then cn = ( f , q5n). ProoJ Just as in the proof of Theorem 1 in Section 7 . 2 , we reason that f ( x ) is continuous. Then

=cl(4l~4m)+."+~n(4n~4m)+...=~m(4m,4m)=~m1l4ml1~~ so that (7.53) holds. The operations on the series are justified by the uniform convergence. If the system is orthonormal, then I)4,(I= 1 , so that c, = ( f , &).

500

I

Advanced Calculus, Fifth Edition

COROLLARY If under the hypotheses of Theorem 8,

and both series converge uniformly over the interval, then c, = ci for n = 1,2, ....

THEOREM 9 Let {+,(x)) be an orthogonal system for the interval a 5 x 5 b and let f ( x ) be piecewise continuous for a _< .x 5 b. For each n the coefficientscl , . . . , cn of the Fourier series of f with respect to {+n(x))are those constants that give the square error 11 f - g (I2 its smallest value, when g ranges over all linear combinations ~ , + ~ ( x ). . pn+n(x). The minimum value of the error is

+ +

I

I

I

COROLLARY Under the hypotheses of Theorem 9, one has so that the series C c

~ ( I + ~ 112

converges. Furthermore, lim ck (I&

k+00

(1 = 0.

The proofs are left as exercises (Problems 5,6,7 following Section 7.13);(7.55) is Bessel's inequality. A crucial question is whether one can assert that En converges to 0,as n + ao. This is equivalent to asking whether f ( x )can be approximated, in the sense of least square error, as closely as desired by a linear combination of a finite number of the functions +,(x). If this is the case, the system { + n ( ~ ) }is termed complete:

DEFINITION An orthogonal system {+,(x)) for the interval a 5 x 5 b is termed complete if for every piecewise continuous function f ( x ) in the interval a 5 x 5 b the minimum square error En = [If - ( ~ .~. . 4c,+,)((* ~ converges to zero as n becomes infinite.

+ +

If the system is complete, then Theorem 9 shows that

l l f 112

= ci1141112

+ . + c~l14n112+ . . . .

This is Parseval's equation. Conversely, if Parseval's equation holds for every piecewise continuous f , then En must converge to 0, and the system {@,(x))is complete. Therefore the validity of Parseval's equation is equivalent to completeness. I f the system { & ( x ) ]is orthonormal, Parseval's equation becomes

llf112=c;+...+c;+.... This is analogous to the vector relations

Chapter 7 Fourier Series and Orthogonal Functions If the orthogonal system {c$,(x)}is complete, then the square error En converges to 0. This does not imply convergence of the Fourier series C c,#,(x) to f ( x ) , although the successive partial sums do approach f ( x ) in the sense of square error. We describe the situation by saying that the series converges in the mean to f ( x ) , and we write

where "L.i.m." stands for "limit in the mean." In general, if functions f ( x ) and f n ( x ) (n = 1 , 2 , . . .) are all piecewise continuous for a 5 x 5 b , we write L.i.m. fn(x) = f ( x ) n-rm

(7.58)

if the sequence 11 f n ( x ) - f (x)ll converges to 0, that is, if

Remark. The following assertions can be proved: (a) if f n ( x ) converges uniformly to f ( x ) for a 5 x 5 b , then f n ( x ) also converges in the mean to f ( x ) , provided that all functions are piecewise continuous; (b) if f , ( x ) converges in the mean to f ( x ) , then f,(x) need not converge uniformly to f ( x ) ; in fact, the sequence f , ( x ) need not converge for a 5 x 5 b ; (c) if f n ( x ) converges to f ( x ) for a 5 x 5 b , but not uniformly, then f,(x) need not converge in the mean to f ( x ) . For proofs, refer to Problem 9 following Section 7.13. DEFINITION An orthogonal system {c$,(x)} for the interval a 5 x 5 b has the uniqueness property if every piecewise continuous function f ( x ) for a 5 x 5 b is uniquely determined by its Fourier coefficients with respect to { @ n ( x ) }that ; is, if f ( x ) and g ( x ) are piecewise continuous for a 5 x 5 b and ( f , 4,,) = ( g , 4,) for all n , then f ( x ) - g ( x ) = O*. This is equivalent to the statement that h ( x ) = 0* is the only piecewise continuous function orthogonal to all the functions Gn(x);the system of orthogonal functions can therefore not be enlarged. THEOREM 10 Let {@,,(x))be a complete system of orthogonal functions for the interval a 5 x 5 b . Then { + n ( x ) )has the uniqueness property.

The proof is left as an exercise (Problem 8). One might expect the converse to hold, that is, that the uniqueness property implies completeness. However, examples can be given to show that this is not the case. Theorem 12 gives more information on this point. THEOREM 11 Let { 4 , ( x ) } be an orthogonal system of continuous functions for the interval a 5 x 5 b and let {c$,,(x)}have the uniqueness property. Let f ( x ) be continuous for a 5 x 5 b and let the Fourier series of f ( x ) with respect to { @ n ( x ) } converge uniformly for a 5 x 5 b . Then the Fourier series converges to f ( x ) .

The proof is left as an exercise (Problem 10 following Section 7.13).

501

502

Advanced Calculus, Fifth Edition

*7.12 SUFFICIENT CONDITIONS FOR COMPLETENESS THEOREM 12 Let {$,(x)) be an orthogonal system of continuous functions for the interval a p x 5 b. Let the following two properties hold: (a) {@,(x)}has the uniqueness property; (b) for some positive integer k, the Fourier series of g(x) with respect to {4,(x)}is uniformly convergent for every g ( x ) having continuous derivatives through the kth order for a 5 x 5 b and such that g(a) = gl(a) = . . . = g(k'(a) = 0, g(b) = gl(b) = . . . = g(k'(b) = 0. Then the system {4,,(x)}is complete.

Proof. Let f ( x )be piecewise continuous for a 5 x 5 b. We must then show that, given E > 0, a linear combination c ,@l ( x )+ . . . + c,$, ( x ) = + ( x ) can be found such that I( f - $11 < E . For simplicity we assume the integer k of assumption (b) to be 2, as this case is typical. The construction of the function +(x) proceeds in several stages. We first determine a continuous function F ( x ) such that 11 F - f 11 < and F ( x )= 0 when a 5 x 5 a 6 , b - 6 p x 5 b for a proper choice of 6 > 0. This is suggested graphically in Fig. 7.14. We denote by f i ( x ) the piecewise continuous function that coincides with f ( x ) except for a p x p a 26 and for b - 26 p x p b, where f i ( x ) is identically zero. We now bridge the jumps in f l ( x ) by line segments. At each jump the line joins [xo- 6, fi(xo - 6 ) ] to [xo+ 6, fl(xo a)]. The function F ( x ) coincides with fi(x) except between xo - 6 and xo + 6, where it follows the straight line. Accordingly, F ( x ) is a continuous function, and 11 F - f 1l2 is a sum of a finite number K of integrals of form

it

+

+

+

plus two integrals of [f (x)I2from a to a +6 and from b - 6 to b, where F = 0. Since f ( x )is piecewise continuous, If ( x ) ( 5 M for some constant M. By its construction, I F(x)l 5 M also. Hence I F ( x ) - f (x)l 5 2M for every x , and

@

Figure 7.14 Approximation of the piecewise continuous function f (x) by a continuous function F ( x ) and by a smooth function G(x).

503

Chapter 7 Fourier Series and Orthogonal Functions Adding the expressions for all K jumps and for the two ends, we find As 6 + 0, the expression on the right approaches 0. Accordingly, for 6 sufficiently small, [IF- f 11 c ~1 c . We next choose a function G(x) having a continuous first derivative for a _< xsb,suchthatIIG-FII < ~ t a n d ~ ( x ) = 0 f o r a ( x ( a + ~ 6 a n d f o r b - ~ 6 ( x s b . We define F(x) to be identically 0 outside the interval a 5 x 5 b and then set

The constant h will be specified later. The function G(x) is then defined for all x and G(x) = 0 for x 5 a $6, and for x p b - i 6 , as required. By the fundamental theorem of calculus [Sections 4.1 and 4.91,

+

1 G ' ( ~ I= ) -[F(xl h ) - F(xl - h)]. 2h Accordingly, G(x) has a continuous derivative for all x. Now

+

G ( x ~) F(xl) = -

F ( x ) ~ x- F(xl) = -

and therefore

We apply the Schwarz inequality (7.48) to the integral on the right-hand side, with f (x) replaced by F(x) - F(xl) and g(x) by 1. Accordingly,

and [ G ( x ~) F(X~)]'d~

If we set xl = u, x = u - v in the double integral on the right-hand side, it becomes (cf. Section 4.6)

since the region of integration in the uv-plane is the rectangle a Iu 5 b, -h Iv 5 h. If we write H(v) for the inner integral, then H(v) is a continuous function of v (Section 4.1 I), and H(0) = 0. Now

d

504 # Advanced Calculus, Fifth Edition where -h < v* < h, b the Mean Value theorem. As h -+ 0, H ( u * ) must approach O; hence 11 G - F)1< Z E for h sufficiently small. Next we construct a function g(x) such that IIg -G11 < $ E , while g has continuous first and second derivatives and g(x) = 0 for a 5 x 5 a $6 and forb - $6 5 x _< b. We need only repeat the averaging process of the preceding paragraph:

Y

+

The other steps can be repeated, and the desired inequality is obtained by appropriate choice of p. Since

and G has a continuous derivative, g(x) has continuous first and second derivatives for all x . Finally, we construct a linear combination +(x) = cl 41(x) cn4n(x) such that Ilg - I)11 < t . Since g(x) satisfies all the conditions of assumption (b) in the theorem, the Fourier series of g(x) is uniformly convergent. By (a) and Theorem 11 the series converges to g(x). By the remark preceding Theorem 10 the partial sums of this series also converge in the mean to g(x). Accordingly, a partial sum Sn(x) = +(x) can be chosen such that llg - 11 < $ E . The function +(XI is precisely the linear combination sought, for by (7.47),

+

+

+

Ilf - @ I 1

Ilf - F + F - G + G - g + g - $ 1 1 5 Ilf - Fll + llF - Gll + IlG - gll + llg - @I1

=

Remark. The preceding proof shows that the assumption (a) can be omitted if one replaces "uniformly convergent" in (b) by "convergent in the mean to g(x)."

THEOREM 13 The trigonometric system 1, cos x , sin x , is complete in the interval -x 5 x 5 x.

. . . , cos n x , sin n x , . . .

Prooj The hypotheses (a) and (b) of Theorem 12 are satisfied by virtue of the preceding Theorems 3 and 5. Accordingly, the system is complete.

*7.13 INTEGRATIONAND DIFFERENTLATION OF FOURIER SERIES THEOREM 14 An orthogonal system {#,(x)), for the interval a 5 x 5 b, is complete if and only if for every two functions f (x) and g(x) that are piecewise continuous for a 5 x 5 b, one has

Chapter 7 Fourier Series and Orthogonal Functions where c,, c; are the Fourier coefficients of f ( x ) and g(x), respectively, with respect to { A ( x ) ) .

Remark. Equation (7.59)is termed the second form of Parseval's equation. When the system is orthonormal, it reduces to the equation which is analogous to the equations for vectors. Proof of Theorem 14. If (7.59)holds, then by taking g = f , one concludes that Parseval's equation (7.57) holds; hence the system is complete. Conversely, let { & ( x ) ) be complete. Then the algebraic identity AB = :{(A B)' - ( A - B ) ~ ) gives the equation

+

Parseval's equation for f

+ g and f - g now gives

On adding the two series we obtain (7.59).

THEOREM 15 Let { & ( x ) ) be a complete orthonormal system for the interval a 5 x 5 b. Let f ( x )be piecewise continuous for a 5 x 5 b and let g(x) be piecewise continuous for xl 5 x 5 x2, where a (xl < x2 5 b. Let C c,$,,(x) be the Fourier . series of f ( x ) with respect to { @ , ( x ) )Then

Remark. The theorem asserts that the integral on the left-hand side can be computed by term-by-term integration of the series C cng(x)q5,(x).This is remarkable since there is no assumption of convergence-let alone uniform convergence--of the series before integration. Proof of Theorem 15. We extend the definition of g(x) to the whole of the interval a 5 x 5 b by setting g(x) = 0 for a 5 x 5 xl and for x2 I :x 5 b. Then Parseval's equation (7.59)gives

IIA 1"; = ( 8 ,L)=

J

a

so that Eq. (7.60)is valid.

b

g ( x ) ~ ( x ) d= x

1;

g(x)L(x)dx.

r!

505

aalE

506

Advanced Calculus, Fifth Edition If we choose g(x)

COROLLARY

4

E

1 for X I 5 x 5 x2, we obtain the following result:

Under the hypotheses of Theorem 15,

that is, term-by-term integration is permissible for every Fourier series with respect to a complete orthogonal system {$,(x)). Although term-by-term integration causes no difficulties, term-by-term differen. . . (sin n x l n ) . . . tiation calls for great caution. For example, the series sin x converges for all x , but the differentiated series cosx . . . cos nx . . . diverges for all x . Differentiation multiplies the nth term by n, which interferes with convergence; integration divides the nth term by n, which aids convergence. The safest rule to follow is that of Theorem 33 of Section 6.14: Term-by-term differentiation of a convergent Fourier series is allowed, provided that the differentiated series converges uniformly in the interval considered.

+ + + +

+

+

PROBLEMS 1. Let&(x)=sinnx(n = 1,2, ...). a) Show that the functions ( & ( x ) ) form an orthogonal system in the interval 0 5 x 5 n. b) Show that the functions q5n(x)have the uniqueness property. [Hint: Let f (x) be a function orthogonal to all the Let F(x) be the odd function coinciding with f (x) for 0 < x 5 n. Show that all Fourier coefficients of F(x) are 0.1 c) Show that the Fourier sine series off (x) is uniformly convergentfor every f (x)having continuous first and second derivatives for 0 5 x 5 n and such that f (0) = f (n) = 0. d) Show that {&(x)} is a complete system for the interval 0 5 x 5 n . 2. Cany out the steps (a), (b), (c), and (d) of Problem 1 for the functions &(x) = cosnx (n = 0, 1,2, . . .). Show that the condition f (0) = f ( n ) = 0 IS not needed for (c). 3. Prove the validity of (7.44) for functions f (x), g(x) that are piecewise continuous for asxsb. 4. Verify the correctness of the Schwarz inequality (7.48) and the Minkowski inequality (7.49) for f (x) = x and g(x) = ex in the interval 0 5 x 5 1. 5. Prove the corollary to Theorem 8 [see the proof of the corollary to Theorem 1, Section7.21. 6. Prove Theorem 9 [cf. Problem 7 following Section 7.41. 7. Prove the corollary to Theorem 9 [cf. the proof of the corollary to Theorem 2, Section 7.41. 8. Prove Theorem 10. [Hint: Show by Parseval's equation (7.57) that if (h. 4,) = 0 for all n, then 11 h 11 = 0.1 9. a) Let the functions fn(x) be continuous for a 5 x 5 b and let the sequence fn(x) confn(x) = f (x). verge uniformly to f (x) for a 5 x (b. Prove that L.i.m.,,, b) Prove that the sequence cosnx converges to 0 in the mean for 0 5 x 5 n but does not converge for x = n. Prove that the sequence converges for 0 5 x 5 in but not uniformly.

Chapter 7

Fourier Series and Orthogonal Functions

507

c) Let f n ( x ) = O f o r O ~ x ~ l / n , = n f o r l / n ~ x ~ 2 / n , = O f o r 2 / n ~ x ~ 1 . S h o w that the sequence converges to 0 for 0 5 x 5 1 but not uniformly and that the sequence does not converge in the mean to 0. 10. Prove Theorem 11 [cf. the proof of Theorem 4 in Section 7.71. 11. a) Show that the functions 4,(x) = sin(2n + l)x, n = 0, 1,2, . . . form a complete orthogonal system on the interval I: 0 5 x 5 n/2. [Hint: Verify orthogonality directly. The functions all satisfy the symmetry conditions:

and one verifies that they are the only members of the sequence 1, cosx, sinx, i cos 2 x , sin 2 ,. . . which satisfy (*). Show that if g(x) is defined on I, gl'(x) is continuous on I and g(0) = g(n/2) = gl(0) = gf(n/2) = 0, then g(x) has an extension f (x) of period 2 n which satisfies (*) and is piecewise very smooth, so that its Fourier series converges uniformly to f for all x. As in Problem 5 following Section 7.5, the Fourier series of f contains only terms in sin x, sin 3x, . . . Now apply Theorem 12.1 b) Show that the functions (cos (2n l)x), n = 0, 1 , 2 , . . . form a complete orthogonal system on the interval 0 5 x 5 n/2. By a linear substitution x = a t B, cr # 0, an orthogonal system {&n(x)},a 5 x 5 b, becomes a system (@,(t)} on a new interval with endpoints (a - B)/cY,(b - B)/a. Show that {Ilr,(t)}is orthogonal and is complete, if {&(x)} is complete. Use the indicated substitution as in Problem 12 to obtain a new complete orthogonal system from the given system {q5n(x)]. a) #,(x)=cosnx, n = 0 , 1 , 2,..., O s x i n , x = 2 t b) q5,, as in (a), x = (n/2) - t c) &,(x) = sinnx, n = 1 , 2 , .. ., 0 5 x 5 n , x = t +(n/2) In Problem 12 let the substitution be nonlinear: x = g(t), where g and g' are continuous k, gl(t) # 0, with g(h) = a and g(k) = b or g(h) = b and g(k) = a. for h 5 t Show that the new system @,(t) = q5n(g(t))lgf(t)l'/2 is orthogonal on the interval with endpoints h, k and is complete if (&,(x)] is complete [One says here that the functions @,,(g(t)) are orthogonal with respect to the weight function w ( t ) = Igl(t)l (see Section 7.15).] The choice g(t) = cos-' t (principal value) for 0 5 t 5 1 or for -1 5 t 5 1 does not meet the conditions of Problem 14 because of the discontinuities of gl(t) at f1. However, the resulting improper integrals converge (Section 4.1) and the conclusions remain valid. Apply this substitution to each of the following cases and give the new orthogonal system: a)cosnx, O I x 5 1 r , n = 0 , 1 , 2 ,..., - 1 5 t 5 1 b) sinnx, O s x i n , n = l , 2 ,..., - 1 s t s l c) sin(2n+ l)x, 0 5 x 5 n/2, n = 0 , 1 , 2,..., 0 5 t 5 1 d) cos(2n+ l)x, O 5 x 5 n / 2 , n = 0 , 1 , 2,..., 0 5 t 5 1

+

12.

13.

14.

15.

Remark 1

+

'

For (c) and (d) see Problem 11.

Remark 2 By trigonometry, for n = 0, 1 , 2 , . . . cos (n cos-' t) is a polynomial in t : cos(2cos-' t ) = 2t2 - 1, for example. Hence from (a) one obtains a system of polynomials Tn(t) which are orthogonal for -1 5 t 5 1 with respect to a weight function.

508

Advanced Calculus, Fifth Edition A similar statement applies to (b),( c )and (d),since sin (ncos-' t ) equals a polynomial Un(r) times sinx = ( 1 - t2)lI2.The polynomials Tn(t)and Un(t)are the Chebyshev polynomials o f first and second kinds, respectively. They are special cases of the Jacobi polynomials p,(a9')(x)defined in Section 7.15: a! = /? = - Z 1 for the first kind, a = B = 1 for the second kind.

I

In Theorem 13 we exhibited a key example of a complete orthogonal system: the trigonometric system of sines and cosines. In Problems 1,2, and 11 to 15 following Section 7.13 we provided further examples, all related to the system of sines and cosines. In this section we present an example of a different kind, that of the Legendre polynomials; in the following section, other examples are considered. The Legendre polynomials Pn(x)(n = 0 , 1 , 2 , . . .) can be defined by the formula of Rodrigues:

Thus

THEOREM 16 Pn(x) is a polynomial of degree n. P n ( x )is an odd function or an even function according to whether n is odd or even. The following identities hold forn = 1,2, ... :

Proof. The statements concerning the degree and the odd or even character can be verified from the definition. We recall the Leibnitz Rule: ( u . v)(") = u d n ) n u ' v ( n - l ) + . . . + (:)u(')v("-" . . ., where (:) is the general binomial coefficient n ! / [ r!(n - r ) ! ] .To prove identity (a), we let u = x 2 - 1 . Then

+

+

Chapter 7 Fourier Series and Orthogonal Functions by the Leibnitz Rule. To prove identity (b), we write Pn(x) in two different ways:

If the second equation is subtracted from twice the first, (b) is obtained. It will be seen that the identities (a) and (b), plus the fact that Po(x) = 1, are sufficient to establish all other properties needed, without reference to the definition (7.62). THEOREM 17 relations:

The Legendre polynomials satisfy the following identities and

i) IPn(x)l 5 1 (1x1 5 j) Pn(x)Prn(x)dx = 0 (n # m),

11,

k) f f l [ ~ n ( x ) 1 2 d x=

2 -9

1) xn can be expressed as a linear combination of Po(x), . . . , Pn(x). The proofs are left to the exercises (Problems 2 to 9 following Section 7.15). Identity (d) is the differential equation satisfied by each Pn(x); its importance will be seen in Chapter 10. Identity (e) is known as the recursion formula for Legendre polynomials. It expresses each polynomial in terms of the two preceding members of the sequence; hence knowing that Po = 1 and PI = x, one can successively determine P2, P3, . . . from (e) alone. By virtue of (j) and (k) we can state: THEOREM 18 The Legendre polynomials Pn(x)(n = 0, 1 , 2 , . . .) form an orx < 1, and thogonal system for the interval - 1 I

510

Advanced Calculus, Fifth Edition Accordingly, we can form the Fourier series of an arbitrafj; h'nctiofi f (x ypiecewise continuous for -1 5 x 5 1, with respect to the system {Pn(x)}:

It will be seen that this Fourier-Legendre series behaves in essentially the same way as the trigonometric Fourier series. THEOREM 19 The system of Legendre polynomials in the interval has the uniqueness property.

-1 5 x ( 1

The proof of Theorem 3 (Section 7.7) can be repeated with slight modifications. By part (I) of Theorem 17, every polynomial in x is expressible as a finite linear combination of Legendre polynomials; hence it is sufficient to construct a polynomial pulse-function P(x). One immediately verifies that the polynomial

has the same properties as the function P(x) used in Section 7.7, so that the proof can be completed in the same way. THEOREM 20 If f (x) is very smooth for - 1 5 x series of f (x) converges uniformly for - 1 (x 5 1. Proof.

( 1, then

By use of (c) and integration by parts we find for n

Cn

=

2n

+

the Fourier-Legendre

>2

(' f (x)Pn(x) dx

1

1

= n2

-

;

I

, -

f1(x)(pn+~- ~ n - l ) d x .

2

-1

Because of (f), the first term is 0. Integrating by parts again, we find

1 --

(f ", Pnt2) 4n + 6

1

(f

"3

n )-

1

(f

" 9

Pn)

1 + 4n-2 (f

Accordingly, since 4n - 2 is the smallest denominator,

1

5(Ilf"ll 4n - 2

. IIPn+2l1 + 211f''ll . IIPnII + IIfl'll . IIPn-211),

"9

Pn-2).

Chapter 7 Fourier Series and Orthogonal Functions by the Schwarz inequality (7.46). By (7.65),

By part (i) of Theorem 17 the Weierstrass M-test is applicable to the FourierLegendre series of f (x): for n ? 2,

The series C Mn converges by the integral test. Therefore the Fourier-Legendre a series converges uniformly for - 1 5 x 5 1. COROLLARY If f (x) is very smooth for - 1 (x 5 1, then the Fourier-Legendre series of f (x) converges uniformly to f (x) for - 1 < x 5 1 . This is a consequence of Theorems 1l , 1 9 , and 20. Remark. No condition of periodicity or other condition at x = f1 is imposed on f (x), as in the analogous theorem (Theorem 5 ) for trigonometric series. This is because of a symmetry of the Legendre polynomials, somewhat like that of the functions cos nx (Problem 2 following Section 7.13). THEOREM 21 The Legendre polynomials form a complete orthogonal system a for the interval - 1 < x < 1. This follows at once from the two preceding theorems, by virtue of the general Theorem 12 of Section 7.12. Remark. Given a sequence fn(x) of functions continuous for a < x 5 b, no finite number of which are linearly dependent, one can construct linear combinations = f l , @2 = al fl + a2f2, . . . such that the functions {+n(x)) form an orthogonal system in the interval a 5 x 5 b. This is carried out by the Gram-Schmidt orthogonalization process, described in Section 1.14. If we choose the sequence fn(x) to be 1, x , x 2 , . . . and the interval to be - 1 p x p 1, the functions +,(x) turn out to be constants times the Legendre polynomials. The case of a function f (x) that is piecewise very smooth is covered as for Fourier series by studying a particular jump function s(x). We do not carry through the details here but merely state the result: Just as does the Fourier series, the Fourier-Legendre series converges uniformly to the function in each closed interval containing no discontinuity and converges at each jump discontinuity to the halfway value

It is to be emphasized that there is no peculiar behavior at the points x = f1; here the series converges to the function (provided that the function remains continuous).

511

Advanced Calculus, Fifth Edition For a more complete treatment the reader is referred to the book by Jackson listed at the end of the chapter.

We consider first a special case of the Bessel functions. The Bessel function o f j r s t kind of order 0 is defined by the equation x2 J0(x) = 1 - 22

x4 ++-.+(24(2!)2

x 2n I)"-+... 2 (n!)

.

(7.67)

The ratio test shows at once that this series converges for all values of x. The series is somewhat suggestive of that for cos x , and as the graph of Fig. 7.15 shows, Jo(x)does resemble a trigonometric function. In particular, Jo(x)has injnitely many roots. The positive roots will be denoted, in increasing order, by A , , A 2 , . . . , A k , . . . . It can be shown that as k increases, these roots approach more and more closely the spacing at intervals of n of the roots of sin x or cos x . The function Jo(x) deviates from the pattern of the trigonometric functions in that lim Jo(x) = 0. X

'DO

The oscillation represented by Jo(x)is "damped." The rate of damping is shown by the following "asymptotic formula":

where r ( x ) is bounded:

Figure 7.15

The Bessel function Jo(x).

Chapter 7

Fourier Series and Orthogonal Functions

In other words, the function

+

differs from the function sin ( x in)by a quantity const. r ( x ) / x , which approaches zero as x becomes infinite. Because of the orthogonality of the trigonometric functions, it is not surprising that there is a corresponding orthogonality relation for the functions , 6 J o ( A n x ) :

The condition ( 7 . 7 0 ) is often described by stating that the functions Jo(k,,x) are orthogonal with respect to the weight function p ( x ) = x :

The functions &Jo(A,x) can now be made the basis of a Fourier-Bessel series. For simplicity we write the function being expanded as ,6f ( x ) , so that the series for f ( x ) is

Since it can be shown that

the series for f ( x ) is given by

This series is usually referred to as the Fourier-Bessel series of order 0 of f ( x ) . Because of the close relationship between the functions f i Jo(Anx)and trigonometric functions, it is natural to expect convergence theorems analogous to those for Fourier series. It can indeed be shown that the system {&Jo(Anx)} is complete for the interval 0 5 x 5 1 . Hence by Theorem 1 1 of Section 7 . 1 1 the Fourier-Bessel series of f ( x ) converges to f ( x ) whenever the series is uniformly convergent. The series can be shown to be uniformly convergent i f f ( x ) has, for example, a continuous second derivative and f ( 1 ) = 0. The peculiar requirement at x = 1 is due to the fact that when x = 1, every function Jo(A,,x)is 0. If f ( x ) is piecewise very smooth, an analysis of jump functions again shows that the series is uniformly convergent to f ( x ) except near discontinuity points, where the series converges to the average of left- and right-hand limits. For details on the theory of such expansions, the reader is referred to Chapter XVIII of the treatise of Watson listed at the end of the chapter.

513

514

Advanced Calculus, Fifth Edition The general Bessel function offirst kind and order m is denoted by Jm(x)and is defined by the series:

The function r ( x ) was defined for all x except 0, -1, -2, -3, . . . in Section 6.25. The definition (7.72) therefore fails if m is a negative integer. However, it can be used for these values if we interpret 1/ r ( k ) to be 0 for k = 0, - 1, -2, . . . . If xm is factored out of the series in (7.72), the resulting power series can be shown to converge for all x . However, if m is not an integer, the factor xm gives difficulty for negative x and possibly for x = 0. Hence we shall concentrate on the functions Jm(x)for x 2 0. These functions are continuous for x > 0 and, if m 2 0, for x 2 0. We can formulate a theorem analogous to Theorem 17 for these Bessel functions: THEOREM 22 The Bessel functions of first kind satisfy the following identities and relations for x > 0: a) 2 J k ( x )= Jm-l(x) - Jm+l(x).

b) C)

~ m ( x= ) Jm-l(x)

+

+ Jm+l(x).

J m - l ( ~= ) J ~ ( x ) ;J,(X), or X ~ J , - ~ ( X= ) ;di ; ; ( ~ m J m ( ~ ) ) .

d d) Jm+l(x)= Jm(x)- Jh(x), or -xPm J,,,+l(x) = E(x-m Jm(x)). e) x2J ~ ( x+)X J;(x) ( x 2- m2)J,(x) = 0 (differential equation of the Bessel functions).

+

I)

$[J;J, (cx)]+ (c2+ T1 -)4rn2 J ; J ~ ( c x ) = 0, c = const > 0.

g) ( a 2 - B 2 ) G ~ J m ( a ~ ) J m ( B ~ ) d ~ = B J m ( a ) J A ( B ) - a J m ( B ) J A ( a ) form , 20, a > o , B>O.

h) ( a 2 - B 2 ) x x ~ m ( a x ) ~ m t B x ) = d xaJm(B)Jm+~(a) - BJrn(a)Jrn+~(B), for m>O, a > 0 , B > 0 . 1

i) 2aJ0 x J i ( a x ) d x = J m ( a ) J m + ~ ( a ) + a [ J m ( a ) J A + ~ ( a ) - Jm+~(a)JA(a)l, for m >_ 0, a > 0. j ) The equation Jm(x) = 0 has infinitely many positive roots Aml,Am2,. . . , where A,, < Am2 < . . . < Amn < + .. , Am,, -+ 00 as n -+ CQ, and each root is simple (that is, JA(x) # 0 at each root). k) The equations Jm(x)= 0 and Jm+l(x)= 0 have no common positive roots. 1) Between two successive positive roots of Jm(x)= 0 there is one and only one root of Jm_l(x)= 0 and one and only one root of Jm+1(x)= 0. m) For each fixed m 2 0 the functions ,hJm(Amnx),n = 1 , 2 , 3 , . . . , form an orthogonal system on the 1nterval05 x 5 1 . I" n) S,1 xJ;(Amnx)dx = $ [ ~ ~ + ~ ( A m 1 1 ) ] ~ .

--.

11

0) Form = 0 , 1 , 2 , ..., Jm(x)= $ J ~ c o s ( x s i n O - m 6 ) d 0 . p) Form = 0 , 1 , 2 ,..., IJm(x)J5 1 forx 2 0.

Chapter 7 Fourier Series and Orthogonal Functions The proofs are left to Problem 10. It can be shown that the orthogonal system in part (m) is complete, so that there is a corresponding Fourier-Bessel series of order m. This series is uniformly convergent for every function f i f (x) such that f (x) has a continuous second derivative in the interval and f (1) = 0. For proofs, see Chapter XVIII of the treatise by Watson. On page 406 of the same book it is shown that assertion (p) can be extended to all m 2 0. The Bessel functions of the second kind are the functions Ym(x) (m = 0, 1,2, . . .) defined as follows: 1

Ym(x) = -

-c

(-l)k

k=O

m-1 --1

((im+2k

k!(m + k)!

X

+ 2y -

bm+k

- b,

(7.73)

(m - k - l)! k! '

(;)-m+x

k=O

+

(2log

+ +

R - ' , bo = 0, and y is defined in Eq. (6.88). These where bk = 1 2-I . functions satisfy the same differential equations (e) as the functions of first kind and have other analogous properties. The complex-valued functions

are called Bessel functions of the third kind or Hankelfunctions. For details, see the book by Watson. Tables and graphs of both Legendre polynomials and Bessel functions are available. The book by Jahnke and Emde listed at the end of the chapter provides such data and also a convenient summary of properties of the functions. We mention, without further discussion, several other important systems of orthogonal functions: The Jacobi polynomials. For each a! > - 1, B > - 1, the system of polynomials (Pjp.@'(x)](n = 0, 1,2, . . .) is defined by the equations

+

form a complete orthogonal The functions ~ $ ~ ( x=) (1 - x)fa(lj x)i@~fp.@)(x) system in the interval - 1 5 x 5 1. When a = , ! I = 0, the functions C$,(x) reduce to the Legendre polynomials. The Hermite polynomials. The system of polynomials (H,(x)} is defined by the equations:

They are orthogonal over the injinite interval -00 < x < cm with respect to the weight function e- f "'.

515

I

516

Advanced Calculus, Fifth Edition The Laguerrepolynornials.

For each cr > - 1 the system of polynomials ( L ~ ' ( x ) ]

is defined by the equations:

They are orthogonal over the injinite interval 0 5 x < oo with respect to the weight function xffe-I. For the significance of the infinite interval and further information on these functions w e refer to the book of Jackson listed at the end of the chapter. All of these functions arise in a natural way in physical problems. This is illustrated in Chapter 10, where it will be shown that each of a large class of physical problems automatically provides a complete system of orthogonal functions.

PROBLEMS 1. Graph Po(x), Pl(x), . . . , P ~ ( x ) . 2. Prove the following parts of Theorem 17: (i) part (c), (ii) part (d), (iii) part (e). Each of these can be deduced from (7.63) alone. 3. Prove part (f) of Theorem 17 by induction, with the aid of the recursion formula (e). 4. Prove part (g) of Theorem 17 by squaring (a) and (b) of (7.63) and eliminating the terms in P,-, from the equations obtained. 5. Prove part (h) of Theorem 17. [Hint: Show by induction that (g) gives the chain of incquahtles:

6. Prove part (i) of Theorem 17 as a consequence of part (h). 7. Prove the orthogonality condition (j). [Hint: For every very smooth function f (x) for - 1 5 x 5 1, let f *(x)= [(I - x2)f '(x)]'. Prove by integration by parts that ( f *, g) (f, g*) = 0. Take f = P,,(x), g = Pm(x)and use the differential equation (d) to replace f * by -n(n + 1)f and g* by -m(m + l)g, and conclude that (f, g)[m(m + 1) n(n + I)] = 0.1 8. Prove part (k) of Theorem 17. [Hint: Use the recursion formula (e) to express Pn in terms of and and use the orthogonality condition to show that (P,, Pn) = [(2n - l)/n](x P,, Pn-,). Apply the recursion formula to x PI, and use orthogonality to P,,-1)[(2n - 1)/(2n I)]. Now prove by induction that show that (P,. PI,) = (Pn. P n ) = 2 / ( 2 n + 11.1 9. Prove part (1) of Theorem 17 by induction, using the fact that P,,(x) is a polynomial of nth degree. 10. (a). . . (p). Prove the corresponding parts of Theorem 22 with the aid of the following hints: For (a) and (b), use (7.72). For (c) and (d), use (a) and (b). For (e), use (a) through (d). For (f), use (e). For (g), let u(x) = fiJ,,(crx), v(x) = &Jnl(px). Take c = cr and then c = p in (f) to obtain equations u"(x) + . . . = 0, vU(x)+ . . . = 0. Multiply the first equation by -v, multiply the second by u, and add. Then integrate the result from b to 1, noting that uv" - uu" = (uv' - vu')'. Replace u and v by thcir expressions in terms of Jm and finally let b -+ 0 +.For (h), use (g) and (d). For (i), divide both sides of (h) by cr - ,6 and then, with cr fixed, let P -+ a . For (j), note that if Jm(a) = 0 and JA(cr) = 0, then by (g) x Jm(ax)J m ( ~ x ) d = x 0 for all p #a. Let ,6 + cr to conclude that J,,,(crx) = 0

+

I

1 I

I I

1

[d

Chapter 7 Fourier Series and Orthogonal Functions for 0 5 x 5 1, and this is impossible. Hence each root is simple. Deduce from this that there are only finitely many roots in each finite interval, so that the roots can be numbered in increasing order as stated and, if the sequencc is infinite, ~t must diverge to oo.If the sequence were to stop, then for example, Jm(x)> 0 for x > xo. Take c = 1 in (f) to conclude that with g(x) = fiJm(x), g(x) > 0 and gl'(x) < 0 for x > xl, for some xl. If g(x) -+ rn as x + co,then by (f), gtl(x) -+ -m as x -+ oo;show that this is impossible, so that g(x) is bounded and that g(x) must have a limit k as x -+ co.Show by (f) that eithcr k > 0 or k = 0 is impossible.Thus the sequenceof roots must be infinite. For (k), use (h). For (I), use (c), (d), and Rolle's theorem. For (m), use (g). For (n), use (i) and (dl. For (o), use induction. Form = 0, replace cos (x sin 8)by 1 - (x2 sin28/2!)+. . . and use the formula 1; sin2m0 d0 = IT 1 . 3 . . . (2rn - 1)/[2 .4.. .2rn]. Show, with the aid of integration by parts, that g,(x) = n-' cos(x sin0 - rn0)dO satisfies gm+l(x) = (m/x)gm(x) - gh(x), which by (d) is the same recursion formula as is satisfied by Jm(x).For (p), use (0). 11. Prove that for m = 1 , 2 , 3 , . . . , J-,(x) = (J,(x). 12. Prove:

:i

a) J;(x) = z s i n x

b) J-; (x) =

&cos x

e) J; (XI =

(9- cos x)

The theory of Sections 7.10 to 7.12 can be generalized with minor changes to functions of several variables; one need only replace the interval by a bounded closed region R and the definite integral by a multiple integral over R. For example, in two dimensions the inner product and norm are defined as follows:

Orthogonal systems are defined as before, and one can consider the corresponding Fourier series. Discussion of discontinuities becomes more involved, and it is sufficient for most purposes to restrict attention to continuous functions. The analogues of Theorems 7 to 11 can then be proved essentially without alteration. The generalization of Theorem 12 can be carried out, but this requires more care. Corresponding to the trigonometric system, one has the following system in two dimensions:

1,

sinx,

cosx,

siny,

cosy,

cos p x sin qy,

sinxcosy ,..., sinpxsinqy,

sin px cos qy,

cos px cos qy, . . . .

(9.79)

517

I

This system can be arranged in a definite order to form a system {#,(x, y ) ] , which is orthogonal and complete for the rectangle R: -n Ix 5 IT, -n 5 y 5 n (see Problem 1 following Section 7.19). One can also write the Fourier series as a "double Fourier series": { a , sin px sin qy

+ bpqcos px sin q y

This amounts to a special sort of rearrangement and grouping of the series; if the series is absolutely convergent, then the reasoning of Section 6.10 shows that the double series has the same sum as the single series in any order. If f (x, y ) has continuous first and second derivatives for all (x, y ) and is periodic in both variables:

then an argument similar to that of Section 7.8 shows that the series is absolutely and uniformly convergent to f (x, y).

*7.17 COMPLEX FORMOF FOURIER SERIES From the identity elx

= COS x

'+ i sin x

(j

=

a)

1

of Section 6.19, one derives the relations COSX

=

elx

+

a

elx

e-lx

2

'

sinx =

- e-lx 2i

(7.81)

A Fourier series

5 + &an 2

cos nx

+ bn sin n x )

n=l

can hence be written in the form

The summation from -oo to oo is understood to mean an addition of two series:

If both series converge, the result is clearly the same as the single series on the left-hand side of (7.82).

>

Chapter 7 Fourier Series and Orthogonal Functions The form (7.82) has various advantages. The coefficients c, can be defined directly in terms of f (x):

for the integral on the right is interpreted as

In

f (x)(cos nx - i sin nx) dx

2lr -=

See Section 8.2 for the theory of such integrals of complex funcii'dns. ~ h en' i= 0, this gives iao; when n > 0, the integral equals $(a, - ib,); and when n < 0, it equals $(a_,, ib-,). One has thus the concise statement,

+

whenever the series converges to f (x). For formal work with Fourier series, and even for computation of coefficients, the series (7.84) provides a considerable simplification.

By a suitable limiting process the Eqs. (7.84) lead to the relations 1

f (x) = 2/27;

g(t)eix' dt.

/0°

-00

,fz / 1

g(t) = -

O0

f (x)eFiXtdx .

(7.85)

-cc

Thus under appropriate hypotheses a function f (x) defined for -cc < x < cc can be represented as a "continuous sum" of sines and cosines (el"' = cos xt i sin xt). The integral representing f (x) is termed the Fourier integral of f ( x ) . The Fourier coefficients o f f (x) in this integral representation are the numbers g(t), which form a new function. Equations (7.85) show that the relation between f and g is nearly symmetrical. Equations (7.85) can also be written in real form as follows:

+

f (I) = a(r) = -

Scc

joa(r)cosxt dr + lo B(t) sinxr dr.

~ ( X ) C O SdX~I ,

-cc

1/-f, (x)sinxr dx. do

~ ( 1= )

(7.86)

This is directly analogous to the real form for Fourier series. The validity of the formulas (7.85) or the equivalent formulas (7.86) can be established if f (x) is piecewise smooth in each finite interval and the integral

519

520

Advanced Calculus, Fifth Edition converges. The Fourier integral must in general be taken as a principal value (Section 6.24). It converges to f (x) wherever f is continuous and to the average of left and right limits at jump discontinuities. For a proof the reader is referred to Chapter 5 of the book by Kaplan listed at the end of the chapter. The Eqs. (7.85) are often rephrased in the language of transforms. If w

f (x)e-'XId x ,

-GO

(7.87)

< t < oo,

then one calls the function F the Fourier transform of the function f and writes

F = +[f].

(7.88)

Equations (7.85) then state that

/-,

w

00

f (x) =

2n

~(t)e'"'d t =

2n

~(-t)e'''

-,

dt.

(7.89)

Thus by interchanging the roles of x and t we see that f is itself the Fourier transform of F ( - t ) / ( 2 ~ ) . The Fourier transform has a number of properties that make it exceptionally useful in applications to differential equations. We state several such properties formally here and refer to the books on Fourier theory listed at the end of the chapter for details. First of all, cP is a linear operator:

Next

This property is crucial for differential equations. Here it is assumed, in particular, that f has continuous derivatives through order k. The result must be modified when discontinuities occur. (See (7.101) for such a rule for Laplace transforms.) If c is a real constant, then @[ f (x

- c)] = e-ic'+[ f],

@[eicxf(x)]= F(t - c ) ,

where F = @[f].

From two functions f (x) and g(x) (under appropriate hypotheses), one forms their convolution f * g = h as the function h(x) such that

Then

Examples of Fourier transforms. x 5 0. Then

Let fi(x) = e-" for x > 0 and fi(x) = 0 for

Chapter 7 Fourier Series and Orthogonal Functions Similarly, if f2(x) = @ for x 5 0 and fi(x) = 0 for x > 0, then

+

Now f ~ ( x ) f2(x) = e-IXI = f (x), and hence by linearity,

+

Since f is the Fourier transform of ( 2 ~ ) - I F ( - t ) = [n(1 t 2 ) ] - ' , we conclude (after interchanging x and t) that 1 Q[g] = e-"I, where g(x) = n(1 +x2)' or by linearity

*7.19 THELAPLACE TRANSFORM AS A SPECIAL CASE OF THE FOURIER TRANSFORM The Laplace transform of a function f (x) is defined (Section 6.25) as where F(s) =

I"

A?[ f ]

f (x) e-Sxdx.

= F, (7.96)

Here f need only be defined for x L 0. However, it will be convenient here to define f (x) to be 0 for negative x. The Laplace transform F is a function of s, defined wherever the integral in (7.96) exists. We now allow s to be complex and write s = a it, where a and t are real. Since f (x) = 0 for x < 0, we can write

+

S__ 00

~ ( s=)

dx =

J_m_ f(x)epux

e - l t X dx.

Thus we see that for each fixed a, F(s) is the Fourier transform of f (x)e-Ox:

It can be shown that if f is piecewise continuous on each finite interval and If (x)l < kebx,x 2 0, for some constants k and b, then for each a > b, the Laplace transform F(s) = F ( a it) is defined for -00 < t < oo.

+

EXAMPLE

Let f (x) = 1 for x = 0 (and f (x) = 0 for x < 0). Then

Here If (x)l < 2eox for x >_ 0, so that we can take b = 0, and dip[ f ] = F, where F(s) = l / s = l / ( a + it) for a > 0. Since the Laplace transform is related to the Fourier transform by (7.97), we can apply the theory of the Fourier transform to Laplace transforms. In particular,

521

522

Advanced Calculus, Fifth Edition if 9 [ f 1 = F(s) = F ( a conditions on f ,

+ it), then Eq. (7.89) allows us to write, under appropriate

f (x)e-"" = 2lT

/

-m

F ( o + it) elxtd t

(where, as for (7.89), the integral is a principal value). We can rewrite this result as follows:

This is the inversion formula for Laplace transforms. It permits us to recover a function f (x) from its Laplace transform F(s). The existence of such a formula, as of the analogous formula for Fourier transforms, shows that we are dealing with one-to-one mappings. For f (x) as in the example, we found F(s) = l / s for a > 0.Hence (with principal values understood)

At x = 0, f has a jump discontinuity, and the integral converges to the expected average value, namely, We can list formally other properties of Laplace transforms, all deducible from the corresponding properties of Fourier transforms: The Laplace transform is a linear operator. Corresponding to the formula

i.

a[f'k']

= (itlk@[f ] ,

one has (7.100)

9 [ f 'k'(x)] = s k 9 [ f ] .

However, this is valid only iff (regarded as 0 for x < 0) and the derivatives occurring have no discontinuities on the infinite x-axis. Iff has continuous derivatives through order n, when regarded as a function on the interval x 2 0, then the formula becomes

where all derivatives are regarded as right-hand derivatives. The formulas (7.92) and (7.93) become 9 [ f (X - c)] = e-"9[ f 1, 3 [ e b xf (x)] = F(s - b),

.

.

s "

(7.102)

c > 0,

where F(s) = 9 [f 1.

(7.103)

The formula (7.95) becomes

9If

* g1 = 2 [ J ' ] 9 [ g 1 ,

'

(7.104)

523

Chapter 7 Fourier Series and Orthogonal Functions where, since f = 0 and g = 0 for x c 0, the convolution (7.94) can be simplified to f g = h, where

*

This automatically makes h(x) = 0 for x c 0. For further information on Laplace transforms, see the books by Churchill, Kaplan, and Widder listed at the end of the chapter.

PROBLEMS 1. a) Prove that the functions (7.79) form an orthogonal system in the rectangle R: -n j x l n , - n s y l r r .

b) Expand f (x , y) = x2 in a double Fourier series in R. 2. Represent the following functions as Fourier integrals:

.

a) .f(x) = 0 , x < O ; f ( ~ ) = e - ~x , 2 0 b) f ( x ) = O , x t O ; f ( x ) = l , 0 5 x 5 1 ; f ( x ) = O , x > 1 3. Show that if f is even, then its Fourier integral reduces to

l Am 60

1 1

i

r,

' n. r '1

60

a(t) cos xt dt,

a(t) =

f (x) cos xt dx

0

and that if f is odd, its Fourier integral reduces to #?(I) sinxt dt.

2 O 0

#?(I) = -

f(x) sin x t dx.

0

4. Find the Fourier transform of each function: a) f ( x ) = l , - 1 5 x 5 1 ; f(x)=Ootherwise b) f (x) = ~ e - ~x 2 , 0; f (x) = 0 otherwise C) f (x) = x2, -1 5 x 5 1; f (x) = 0 otherwise 5. a) Prove the rule (7.92). b) Prove the rule (7.93). 6. It is shown in Section 6.25 that fork = 0, 1,2, . . . , L?[xk] = k!/sk+' for real positive s. a) Show that the result is valid for complex s = a it, provided that a > 0. b) Write out the corresponding inversion formula (7.98) for f (x) = xk. 7. For each of the following choices of f and g, evaluate h = f * g and verify that (7.104) holds. (Throughout, f = 0 and g = 0 for x < 0.) a) f(x) = g(x) = 1 forx 1 0 . b) f (x) = 1 and g(x) = x for x 2 0. C) f (x) = 1 and g(x) = ex for x 2 0.

+

A number of physical problems lead one to attempt to extend the concept of "function" to include new mathematical objects, called generalized functions or distributions, differing markedly from traditional functions. We give here an intuitive introduction to the subject and give references for a general discussion.

Advanced Calculus, Fifth Edition

A simple example is that of density, say for mass distributed along a line, the x-axis. If p(x) is the density (in units of mass per unit length), then p(x)dx should give the total mass in the interval a 5 x 5 b. Now suppose that all our mass is concentrated in a single particle of mass 1, located at the origin. It is tempting to assign a density p(x) such that p(x) dx = 0 if the interval a 5 x 5 b does Jab not contain the origin and such that p(x) dx = 1 if the interval does contain the origin. However, these conditions would make p(x) = 0 except "near the origin," where p(x) must suddenly become infinite so rapidly that the integral of p(x) over each small interval containing the origin is 1. This particular density "function" is the Dirac delta function 6(x). Thus we require:

lab

la

The apparently contradictory nature of these properties led to their rejection as meaningless by mathematicians until recently. Then a number of mathematicians, notably L. Schwartz, developed systematic theories of generalized functions, which satisfy all the desired properties in a rigorous manner. The function 6(x) is such a generalized function. We can arrive at the delta function in another way, which suggests a limit process for obtaining this and other generalized functions. Let g(b) denote the total mass on the interval -m < x 5 b, for a distribution of mass (with finite total mass) on the x-axis. Thus for a continuous density p(x), g(b) = fb; p(x) dx and gl(b) = p(b). Now for the case of a single particle of mass 1 at the origin we have difficulty in defining p(x) but can easily find g(x) for each x. For g(x) is simply the total mass to the left of x; hence for the case of the single particle, g(x) = 0 for x < 0,

g(x) = 1 for x

> 0.

(7.107)

Thus g(x) has a jump discontinuity at x = 0. We call the function g(x) in (7.107) the Heaviside unit function. Now from this g(x) we can again try to obtain the density p(x) as the derivative of g(x). It is clear that gf(x)= 0 for x # 0, and it is reasonable to write g'(0) = oo, because of the jump. Thus we have gl(x) = 0 = S(x) for x # 0,

gf(0)= oo = S(0).

(7.108)

Accordingly, we interpret the "function" 6(x) as the derivative ofthe Heaviside unit function. Now the unit function really has no derivative at x = 0, so we have not completely clarified the meaning of 6(x). To go further, we approximate g(x) by a smooth function g,(x) having a derivative for all x; we choose an 6 > 0 and set 1

1

g,(x) = 1 - -e-'/' for x 2 0, g,(x) = -ex/' for x < 0 (7.109) 2 2 (see Fig. 7.16). The function g, (x) has been chosen so that as 6 + 0+, g, (x) + g(x) (except at x = 0, where g,(x) = for all 6). Furthermore, g,(x) has a derivative g: (x) = -e1 26

-x/c

for x 2 0,

1 g:(x) = -ex'' 26

for x < 0,

(7.110)

Chapter 7 Fourier Series and Orthogonal Functions

Figure 7.16

Smooth approximation of Heaviside unit function.

Figure 7.17

Function approximating the Dirac delta function.

and since g,(x) + g(x) as r + 0+, it is reasonable to consider S(x) as the limit of g:(x) as S += O+. Thus S(x) is considered to be the limit of the "pulse" graphed in Fig. 7.17 (a smooth approximation to g'(x)), as r -+ O+. However, if we do indeed pass to the limit, we again obtain the value 0 for x # 0, the "value" oo for x = 0. It is thus not clear what we have gained. We make headway by, in effect, stopping just short of the limit: We regard S(x) as a function very closely approximated by the pulse of Fig. 7.17; the smaller the 6 , the better the approximation. Thus to evaluate the integral of S(x) we write

1.

G(x) dx = lim .+O+

1.

g:(x) dx.

From (7.110) we easily find, say for a < 0 t b, lim

r+O+

1'

g:(x) dx = lim r+O+

I

+ e-bJc)

= 1,

and hence the integral of S(x) from a to b is 1. Similarly, if a < b < 0 or 0 < a < b, the integral is found to be 0. (The secret here is in the interchange of two limit processes, as a closer examination shows.) The procedure followed here for S(x) can be extended to other generalized functions. In general, it provides a method for assigning derivatives to a large class of functions (including all piecewise continuous functions) not having derivatives

525

526

Advanced Calculus, Fifth Edition

Figure 7.18

Function approximating the derivative S1(x).

in the ordinary sense. A systematic development based on the ideas suggested here is given in the book by Lighthill listed at the end of the chapter. Here we simply list some of the results obtained and confine attention to generalized functions related to S(x). We find that generalized functions can be differentiated arbitrarily often and hence obtain Sr(x),6"(x), . . . . The first derivative 6'(x) can be approximated by the derivative of the pulse of Fig. 7.17; hence it appears as in Fig. 7.18. Similar approximations are obtained for higher derivatives. A simple translation along the x-axis provides us with generalized functions of form 6 ( x - c ) , 6'(x - c ) , . . . for constant c . An ordinary function, say piecewise continuous, can also be regarded as a generalized function. We confine attention here to ordinary functions that are formed of a finite number of pieces having derivatives of all orders-for example, the function f ( x ) = 0 for x < 0, = e-" for x 2 0. If we approximate f by a smooth function f , ( x ) , as we did for g ( x ) above, we are led to the conclusion that f ( x )has a derivative f r ( x ) = 6 ( x ) h ( x ) , where h ( x ) = 0 for x < 0 and h ( x ) = -e-" for x > 0 (the value at 0 being immaterial). Similarly, f " ( x )is found to be 6'(x) - 6 ( x ) f ( x ) .

+

+

527

Chapter 7 Fourier Series and Orthogonal Functions In general, we are led in this way to generalized functions of form

+

f ( x ) a16(x- c1)

4

+ a26(x - c2) + . - .+ blS1(x- d l ) + . . . + elSU(x- p l ) + . . . , (7.112)

that is, of the form ordinary function (of the type described) plus a linear combination of translated delta functions and translated derivatives up to a certain order of the delta function. Terms with 0 coefficient are to be suppressed. We now confine our attention to generalized function of this type. Operations on generalized functions. We add or subtract generalized functions of the type considered in the obvious way. For example, (ex

+ 6 ( x ) )+ ( s i n x + 6 ( x ) + S'(x - 1)) = ex + sinx + 2 6 ( x ) + 6'(x - 1).

&q

We can also multiply such a function by a scalar constant, by multiplying each term by the scalar. Multiplication of arbitrary generalized functions is not defined; in particular, 6 ( x ) . 6 ( x ) is not defined. However, if one of the two factors is an ordinary function f ( x )satisfying certain continuity conditions, then one can multiply f by each term of the other factor. In particular, one defines f ( x ) 8 ( x ) to equal f (O)S(x),provided that f ( x ) is continuous at x = 0; one defines f (x)S1(x)to equal f (0)6'(x) - f '(O)S(x),provided that f and f' are continuous at x = 0 . These rules can be justified by the limit process described before; the latter rule can also be justified by the rule for differentiating products. In general, one is led to the definition:

provided that f , . . . , f

(k)

are continuous at x = c . For example,

f(x)~"(x c ) = f (c)8"(x - c ) - 2 f '(c)6'(x - c )

+ f "(c)S(x - c ) .

One can differentiate a generalized function of the type considered, by differentiating each term: The derivative of ~ 6 ( ~ )-( cx) is ~ 6 ' ~ + " (-x c); the derivative of f ( x ) is the ordinary derivative plus contributions from the jump discontinuities. If c is such a discontinuity, then we add a term

[x-+c+ lim f ( x ) - lim f (.x)]8(x - c). X'c-

Thus if f ( x ) = 0 for x < 0,f ( x ) = cos x for 0 < x < n/2, and f ( x ) = - 1 for x > n / 2 , then f ' ( x ) = g ( x ) 6 ( x ) - 6 ( x - ( n / 2 ) ) ,where g ( x ) = 0 for x < 0, g ( x ) = -sin x for 0 < x < n / 2 , g ( x ) = 0 for x > n / 2 . One can verify that the usual rules for derivatives of sums and products hold. consider here only definite Generalized functions can also be integrated. integrals from a to b (where a may be -oo and b may be o o ) and exclude the case

+

&

528

Advanced Calculus, Fifth Edition where terms of form A G ( ~ ) (-x c) occur, with c = a or b. Then we integrate each term separately, with

lb

S(x - c) dx =

1 ifatctb, 0 otherwise;

From these definitions we can now integrate certain products of the form g(x)S(k'(x - c): If a < c < b, then

provided that g(k'(x) is continuous at c. This rule follows from the previous rules, especially the rule (7.1 13) for multiplication. The rule (7.1 15) is a crucial one and is itself often made the starting point for the theory of generalized functions. It shows that S ( k ' ( ~ - C) acts as a "linear functional," assigning real numbers to sufficiently smooth ordinary functions g(x). Fourier and Laplace transforms of generalized functions. we deduce that 6(k'(x - c ) has a Fourier transform:

From the rule (7.1 15)

In particular, @[6(x)] = 1 (constant function). If we could write 6(x) in terms of its Fourier integral, we would have

This equation states that the constant function 1 has a Fourier transform 2n6(t).One normally de$nes the Fourier transform of 1 to be 2778(t), even though the reasoning leading to the definition is purely formal. The definition has been shown to lead to no inconsistencies and is a useful one. In a similar manner, from (7.1 16), one is led ~ Fourier transform 2 7 ~ i ~ 8 ( ~ ' c). (t to assign to ~ * e - ' ' the The Fourier transform can be regarded as a one-to-one linear mapping of a certain vector space of functions onto a second vector space of functions. The process we are following can be regarded as an extension of the linear mapping to larger vector spaces. The extension is chosen to preserve linearity and one-to-one-ness. There is a similar discussion of the Laplace transform, which we know to be essentially a special case of the Fourier transform. We find

+

It is a remarkable fact that for both Fourier and Laplace transforms, generalized functions permit one to circumvent the awkward rules for transforms of derivatives. One finds:

I

i

Chapter 7 Fourier Series and Orthogonal Functions where f is as in the previous rules, except for a finite number ofjump discontinuities, as allowed for generalized functions. In (7.1 19) and (7.120), f ("(x) is treated as a generalized function so that it will contain terms in delta functions arising from jumps in f or its derivatives. Convolutions can also be defined for generalized functions. If the rule "transform of convolution of two functions equals product of two transforms" is to hold, then we must require:

where g("(x) is a generalized function. For in the case of Laplace transforms, (7.121) gives Lqg(x)

* s ' ~ ' ( x ) =] s k 2 [ g ] ,

in agreement with (7.120). More generally, we are led to

and then, by term-by-term operations (as for multiplication), we can obtain the convolution of two arbitrary generalized functions. The fact that convolution becomes multiplication in "transform language" has also been used as a starting point for the theory of generalized functions. For details, see the book by Mikusinslu listed at the end of the chapter.

PROBLEMS 1. a) Verify the graph of g,(x), shown in Fig. 7.16. b ) Graph g:(x), approximating S(x) [cf. Fig. 7.171. c ) Graph the derivative of the pulse of Fig. 7.17, to approximate St(x)[cf. Fig. 7.181. d) Graph the derivative of the function of Fig. 7.18 to approximate SU(x). 2. Another smooth approximation of the Heaviside unit function is the function h,(x) = I 1(1 S ~ ( X / Efor ) )- x E / ~5 x 5 n c / 2 , = 1 for x > n c / 2 , = 0 for x < -nc/2. a ) Graph h,(x). b ) Graph h:(x) and consider this as an approximation to S(x) for small positive 6. C ) Show that h : ( x ) d x has limit 1 as E + O+. 3. Let f ( x ) = 0 for x < 0 , f ( x ) = 3 - x for x > 0.Let g ( x ) = cosx for - 7 ~ . 5 x 5 x , g ( x ) = 0 otherwise. Evaluate the following: a ) S f l ( x ) 2g1(x) b) f"(x) C ) f ( x ) S ( X - 1) d) R(x)s'(x) e) f (x)gl'(x) f ) j!, f(x)S1(x- 1)dx

+

I!,

+

9)

j)

R ( X ) ~ ~ d' (x X )

f ( x ) * S1(x)

h) 2 [ f " ( x ) l k) f ( x )* f l ' ( x )

i) *[d"'x)l

Suggested References Churchill, Rucl V.,and Brown, James W., Fourier Series and Boundary Value Problems, 5th ed. New York: McGraw-Hill, 1993. Churchill, Rue1 V., Operational Mathematics, 3rd cd. New York: McGraw-Hill, 1972. Franklin, Philip, A Treatise on Advanced Calculus. New York: Dover, 1964.

529

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Advanced Calculus, Fifth Edition

.

Franklin, Philip, Fourier Methods. New York: McGraw-Hill, 1949. Jackson, Dunham, Fourier Series and Orthogonal Polynomials (Carus Mathematical Monographs, No. 6). Menasha, Wisconsin: Mathematical Association of America, 1941. Jahnke, E., and E Emde, Tables of Higher Functions, 6th ed., revised by F. Losch. New York: McGraw-Hill, 1960. Kaplan, Wilfred, Operational Methodsfor Linear Systems. Reading: Mass., Addison-Wesley, 1962. Lighthill, M. J., An Introduction to Fourier Analysis and Generalized Functions. Cambridge: Cambridge University Press, 1958. Mikusiriski, Jan, Operational Calculus, 2nd ed. New York: Pergamon Press, 1983. Rogosinski, Werner, Fourier Series (trans. by H. Cohn and F. Steinhardt). New York: Chelsea Publishing Co., 1950. Szego, Gabor, Orthogonal Polynomials (American Mathematical Society Colloquium Publications, Vol. 23), 4th ed. New York: American Mathematical Society, 1975. Titchmarsh, E. C., Eigenfunction Expansions, 2nd ed. Oxford: Oxforduniversity Press, 1962. Titchmarsh, E. C., Theory of Fourier Integrals, 2nd ed. Oxford: Oxford University Press, 1948. Watson, G. N., Theory of Bessel Functions, 2nd ed. Cambridge: Cambridge University Press, 1966. Whittaker, E. T., and G. N. Watson, Modern Analysis, 4th ed. Cambridge: Cambridge University Press, 1950. Widder, D. V., The Laplace Transform. Princeton: Princeton University Press, 1946. Wiener, Norbert, The Fourier Integral. New York: Dover, 1951. Zygmund, A., Trigonometric Series. Cambridge: Cambridge University Press, 1968.

Functions of a Complex Variable

We assume familiarity with the complex number system. Figure 8.1 reviews standard notations for the complex z-plane, where z = x i y . We shall denote real and imaginary parts as follows:

+

for z = x + i y ,

x=Rez,

y=Imz.

We shall make steady use of all these notations, as well as the analogous notations in a complex w-plane, where w = u i v . For addition and multiplication by real scalars, complex nunibers can be regarded as vectors in the plane. In particular, lzl 1, 1z21, lzl 121 are the sides of a triangle, as in Fig. 8.1, so that one has the familiar triangle inequality

+

+

+

Complex-valuedfuncrions of z. If to each value of the complex number z = x iy , with certain exceptions, there is assigned a value of the complex number w = u + i v , then w is given as a complex-valued function of z, and we write w = f (2). For example,

532

Advanced Calculus, Fifth Edition

z=x+iy

0 = arg z X

". Figure 8.1

I

'Ir-

The complex z-plane.

are such functions. Important functions of this type are

+ . . + an-lZ + an, aozn + . . + a, w= bozm+ . + bm '

polynomials: w = aozn rational functions:

exp z = eZ= eX+'Y= eX(cosy + i sin y),

exponential functions:

,iz

trigonometric functions:

sin z =

- e-iz

2i

eiz + e - i ~

'

cos z =

2 e-Z

'

ez + 2 The definition of the exponential function is motivated by interpreting eZfor z n / n ! ;see Section 6.19. From the complex z as the sum of its power series definition it follows that for real y , eZ

hyperbolic functions:

sinh z =

- e-2

2 '

C O S ~ Z=

-.

e i Y = c o s y + i s i n y and e - i Y = c o s y - i s i n y , so that

-

e'+ ~ e-i~

e i ~ e-i~

sin y =

cosy =

2 2i ' These equations suggest the definitions given above for sin z and cos z. The definitions of the hyperbolic functions are based on the usual definitions for real variables. By the reasoning described, it follows that when z is real (z = x iO), each of the five functions reduces to the familiar real function. For example, ex+'O = e x . The other trigonometric and hyperbolic functions are defined in terms of the sine, cosine, sinh, and cosh in the usual way.

+

It will be convenient to represent paths for line integrals in the complex plane by equations of form

Chapter 8 Functions of a Complex Variable Here t is a real variable and F is a function whose values are complex, so that we are dealing with a complex-valued function of a real variable. Examples are the following functions:

+

+

In (8.2) we can write z = x i y and F ( t ) = f ( t ) ig(t), where f and g are real-valued. Then (8.2) is equivalent to the pair of equations

We can also consider (8.2) as an alternative to the familiar vector function representation of a path. The path z = e" given previously is equivalent to the path x = cos t , y = sin t , and hence its graph is a circle. The calculus can be developed for complex-valued functions oft in strict analogy with the development for real-valued functions. We write lim F ( t ) = c = xo + iyo

t-+$

if, given E > 0, we can choose 6 > 0 so that 1 F ( t ) - c ) < E when 0 < It - t o ) < 6 . It is assumed here that F ( t ) is defined for t sufficiently close to to, but not necessarily for t = to; if F ( t ) is defined only in an interval to < t < 8, the limit is interpreted as a limit to the right and is written lim F ( t ) .

r-+ro+

Limits to the left are defined similarly. Limits as t for real functions. [However,

+ oo or t + -oo are defined as

lim F ( t ) = oo

t+t"

is defined to mean lim IF(t)l = oo;

r+ro

there is no concept of +oo or -00 for complex numbers; see Section 8.14.1 If F ( t ) is defined for cr < t < and to lies in this interval, then F ( t ) is said to be continuous at to if lim F ( t ) = F(to).

r'ro

If F ( t ) is also defined at t = a and lim F ( t ) = F ( a ) ,

r-rlY+

then F ( t ) is said to be continuous to the right at t = a. Continuity to the left and continuity in an interval are defined as for real functions. If F ( t ) = f ( t ) ig(t), then Eq. (8.3) is equivalent to the two equations

+

533

534

,

Advanced Calculus, Fifth Edition

y - 7 ~ 1r s

For Eq. (8.3) signifies that F ( t ) is as close to c as desired, for t sufficiently close to to; by a geometric argument we see that this is equivalent to the requirement that f = Re F be as close to xo as desired and g = Im F be as close to yo as desired, for t sufficiently close to to. Similarly, continuity of F(r) at to is equivalent to continuity of f ( t )and g ( t ) at to. Accordingly, such functions as t 2 it and e" are continuous for all t . Furthermore, the rules for limits of sums, products, and quotients and the analogous continuity theorems must cany over to the complex case. For example, if F l ( t )= f ~ ( t ) i g , ( t ) and F2(t)= f 2 ( t ) ig2(t)are continuous in an interval, then so is

+

+

+

For fi ( t ) ,gl ( t ) ,f 2 ( t ) ,g2(t)must be continuous, so that the real and imaginary pans of Fl ( t ) . F2(t)are continuous, and hence Fl ( t ) . F2(t)is continuous. For the discussion thus far we are to some extent repeating the theory of vector functions. However, we remark that complex multiplication and division have no analogue for vectors. Derivatives and integrals. The derivative of F ( t ) can be defined as for real functions: F(to A t ) - F(t0) Ft(to)= lim

+

At

Ar+0

By taking real and imaginary parts and applying (8.4)we conclude that

+

F1(to>= f '(to) igl(to);

(8.6)

that is, F ( t ) has a derivative at to precisely when f ( t ) ,g ( t ) have derivatives at to, and the derivatives are related by Eq. (8.6).Derivatives to the left or right are defined by requiring that At < 0 or At > 0, respectively, in Eq. (8.5);Eq. (8.6)also applies to these derivatives. The rules for derivative of sum, product, quotient, constant, and constant times function all carry over, and the proofs for real functions can be repeated. We also note the rules

-e,a+b,V = (a + bi)ela+bi'r ( a ,b real). dt

Furthermore, if F 1 ( t )= 0 for a! < t < p, then F ( t ) is identically constant for t < p. The proofs are left as exercises (Problems 9 to 11). Higher derivatives are obtained by repeated differentiation:

(8.8) a!

<

The first derivative of F ( t ) can be thought of as the velocity vector of the point ( x , y ) as it moves on the path x = f ( t ) ,y = g ( t ) ,with t as time. The second derivative can be interpreted as acceleration.

T

Chapter 8 Functions of a Complex Variable The dejinite integral of F ( t ) over an interval a _( t 5 B is defined as a limit of a sum C F(t;) Akt as for real functions. However, again the limit theorem permits us to take real and imaginary parts:

/

If F ( t ) is continuous over the interval, then f ( t ) and g ( t ) are continuous, so the integral exists. More generally, the integral exists if f ( r ) and g ( t ) are piecewise continuous for a ( t 5 p. When f and g are piecewise continuous, we say that F = f ig is piecewise continuous. An indefinite integral of F ( t ) is defined as a function G ( t ) whose derivative is F(t). As in ordinary calculus, we find that if G ( t ) is one indefinite integral, then G ( t ) c provides all indefinite integrals:

+

+

S

+

r:

F ( t )d t = G ( t ) c ,

c being an arbitrary complex constant. If an indefinite integral G of F is known, then it can be used to evaluate definite integrals of F as in calculus:

The proof is left as an exercise (Problem 12). From Eq. (8.9) we can verify the familiar rules for the integral of a sum, the integral of constant times function, the combination of integrals from a to j3 and from /Ito y , and integration by parts. We also have the basic inequality

IIB 1 c JB ~ ( dt t )

IF ( ~ )dtI i M(B - a ) ;

this is valid if a < p, if F ( t ) is, for example, piecewise continuous for a 5 t 5 j3, and if 1 F(r)l 5 M on this interval. The inequality is most easily obtained from the definition of the integral as limit of a sum; for we have, by repeated application of the triangle inequality (8.l),

I

1 ,:*

F ( t l ) Art 5

lF(tl)lA,t 5 M(j3 - a ) ,

and passage to the limit gives (8.1 1). Definite integrals and indefinite integrals are related by the rule

(see Problem 12). EXAMPLE 1

535

536

Advanced Calculus, Fifth Edition

EXAMPLE 2 e(a+bi)r 1 ea+bi - 1 e(a+bl)tdt = - =

a+bio

a+bi

(a

+ bi # 0).

EXAMPLE 3

where p ( t ) is a polynomial of degree n and a is a complex constant, not 0. The equation is established by integration by parts [Problem 8(f)].

PROBLEMS 1. a) Let zl = rl(cos81

+ i sine,),

22

+ i sin82). Show that + 82) + i sin (81 + 8d1.

= r2(cos 82

zl . z2 = rlr2[cos (81

b) Show that e1('l+'2) = elel eLe2[see part (a)].

2. Prove the following identities: a)

eZi

. eZ2 = eZ1+z2[see Problem 1(b)]

b) (eZ)" = enZ (n = 0, f1, f 2 , . . .) c) sin2 z cos2 z = 1 z2) = sin z 1 cos z2 cos z 1 sin 22 d) sin (z e) Re (sin z) = sin x cosh y , Im (sin z) = cos x sinh y f) siniz = i s i n h z , c o s i z =coshz g) e Z = e i , ~ = s i n ? , ~ = c o s Z 3. a) Prove that eZ # 0 for all z. b) Prove that sin z and cos z are 0 only for appropriate real values of z. 4. Represent the following functions graphically: a) w = (1 t) i(1 - t) C) w = e3ir b) w = t4 + i(t2 1) d) = 2e(-1+2i)t ), = te(-l+2i)t f ) w = e-' - iei'

+ +

,

+ +

+

+

5. Find the derivatives of the functions of Problem 4. 6. Graph w = 3e2" and indicate the first and second derivatives graphically for t = n/2, t = n . 7. Integrate the functions of Problem 4 from 0 to 1. 8. Use integration by parts to evaluate each of the following: a) l ( 1 it12 sin t d t b) j tne-a' d t (n = 1 , 2 , . . .)

t

= 0,

+

C)

S t n sinbtdt = $ S tn(eb" - e-bir)dt (n = 1 , 2 , ...)

d) l t n c o s a t d t = R e i t n e i a t d t ( a r e a l , n = 1,2, ...) e) j t n c o s a t c o s b t c o s c t d t (n = 1 , 2 , ...) f ) jp(t)e-'' dt, where p(t) is a polynomial of degree n (Example 3 in the text) 9. Prove Eq. (8.7) by induction (repeated application of rule for differentiation of a product). 10. Prove (8.8) with the aid of (8.6).

537

Chapter 8 Functions of a Complex Variable 11. Prove that if F1(t)= 0,a! < t < /I, then F ( t ) E const for a! < t < /I. 12. a) Prove (8.12) by taking real and imaginary parts. b) Prove (8.10) either directly or as a consequence of (8.12).

FUNCTIONS OF A COMPLEX 8.3 COMPLEX-VALUED VARIABLE LIMITSAND CONTINUITY We return to the general complex-valued function of a complex variable. These functions will be our principal concern for the remainder of this chapter. We write where z = x function

+ iy, w

=u

+ i v , to indicate such a function. An example is the w

= z2 (all z).

Here we can also write:

so that (on taking real and imaginary parts) U

=x

2

v = 2xy.

WY2,

In a similar manner, every complex function w = f (z) is equivalent to a pair of real functions:

of the two real variables x, y. Also from such a pair of real functions, defined on the same set, we obtain a complex function of z. For example, u = x 2 + x y + y 2,

v=xy3

is equivalent to the complex function

w = f(z)=x2+xy+y2+xy3i,

+

+ + +

+

for which f (1 2i) = 1 2 4 8i = 7 8i. The functions eZ,sin z , cos z, sinh z , and cosh z were defined in Section 8.1. For these the corresponding pairs of real functions are as follows: w = ez:

u = ex cosy,

v = ex siny,

w=sinz:

u=sinxcoshy,

v=cosxsinhy,

w=sinhz:

u=sinhxcosy,

v = C O S ~ Xsin y, 8

The proofs are left as exercises (Problem 1 following Section 8.4). In (8.13), each function is defined for all z, that is, for all (x, y).

>

538

Advanced Calculus, Fifth Edition

a z-plane

Figure 8.2

Domain and neighborhood.

In general, we assume w = f (z) to be defined in a domain (open region) D in the z-plane, as suggested in Fig. 8.2. If tois a point of D, we can then find a circular neighborhood /z - zoI < k about zo in D. If f ( z ) is defined in such a neighborhood, except perhaps at zo, then we write lim f (z) = wo ZZ'O

if for every r > 0 we can choose 6 > 0,so that

.

- 2 -

. '*

If f (20) is defined and equals wo and (8.14) holds, then we call f (z) continuous at 20.

+

THEOREM 1 The function w = f (z) is continuous at zo = xo iyo if and only if U(X,y) = Re [f (z)] and v(x, y) = Im [f (z)] are continuous at (XO,yo). Thus w = z2 = x2 - y2 + 2ixy is continuous for all z, since u = x 2 - y2 and v = 2xy are continuous for a11 (x, y). The proof of Theorem 1 is left as an exercise (Problem 5 following Section 8.4). THEOREM 2 The sum, product, and quotient of continuous functions of z are continuous, except for division by zero; a continuous function of a continuous function is continuous. Similarly, if the limits exist, lim [f (z) ZZ'O

+ g(z)l = lim f (z) + lim g(Z), . . . Z'ZI

ZZ'O

These properties are proved as for real variables. (It is assumed in Theorem 2 that the functions are defined in appropriate domains.) It follows from Theorem 2 that polynomials in z are continuous for all z and that each rational function is continuous except where the denominator is zero. From Theorem 1 it follows that eZ= e" cosy + i e x siny is continuous for all z. Hence, by Theorem 2, so also are the functions e i ~- e-iz

sin z =

2i

,iz

'

cos z =

+e-i~

2

539

Chapter 8 Functions of a Complex Variable

f

We write lim f ( z ) = oo if zlim - + z ~ I~ f (z)l = +m;

Z'ZO

that is, if for each real number K there is a positive 6 such that 1 f ( z ) J> K for 0 < Iz-zo I < 6 . Similarly, iff ( z )is defined for lzl > R, for some R, then lim,,, f ( z )= c if for each E > 0 we can choose a number RO such that If ( z ) - c ( < E for (21 > Ro. All these definitions emphasize that there is but one complex number oo and that "approaching m" is equivalent to receding from the origin.

8.4 DERIVATIVES AND DIFFERENTIALS Let w = f ( z ) be given in D and let zo be a point of D. Then w is said to have a derivative f ' ( z o )if lim

A:-0

f

(20

+ A z ) - f (20) = f A2

l(zo>

In appearance this definition is the same as that for functions of a real variable, and it will be seen that the derivative does have the usual properties. However, it will also be shown that if w = f ( z ) has a continuous derivative in a domain D, then f ( z ) has a number of additional properties; in particular, the second derivative f " ( z ) ,third derivative f "'(z), . . . , must also exist in D. The reason for the remarkable consequences of possession of a derivative lies in the fact that the increment Az is allowed to approach zero in any manner. If we restricted Az so that zo Az approached zo along a particular line, then we would obtain a "directional derivative." But here the limit obtained is required to be the same for all directions, so that the "directional derivative" has the same value in all directions. Moreover, zo A z may approach zo in a quite arbitrary manner-for example, along a spiral path. The limit of the ratio A w l A z must be the same for all manners of approach. We say that f ( z ) has a differential d w = c A z at zo if f (zo Az) - f (20) = c A z + r A z , where E depends on A z and is continuous at A z = 0,with value zero when A z = 0.

+

+

+

THEOREM 3 If w = f ( z ) has a differential d w = c A z at zo, then w has a derivative f l ( z o ) = c . Conversely, if w has a derivative at zo, then w has a differential at to:d w = f '(zo)Az. This is proved just as for real functions. We also write A z = d z , as for real variables, so that

From Theorem 3 we see that existence of the derivative f ' ( z o )implies continuity o f f at zo, for

.

I!

540

Advanced Calculus, Fifth Edition THEOREM 4

If wl and w2 are functions of z that have differentials in D, then

If w2 is a differentiable function of w l , and wl is a differentiable function of z, then wherever w 2 [ w (l z ) ]is defined, dw2 - - -d. -~ 2dw1 dz

d w ~ dz'

If w is a differentiable function of z, and z is a differentiable function of the real variable t , then wherever w [ z ( t ) ]is defined, dw - dw dz -- -.dt dz d t '

These rules are proved as in elementary calculus. We can now prove as usual the basic rule:

Furthermore. the derivative of a constant is zero.

PROBLEMS 1. For each of the following, write the given function as two real functions of x and y and determine where the given function is continuous: a) w = (1

+ i)z2

e-L

d ) w = r n g ) w = cosz

w=tanz= f ) w = sinz i) w = coshz

b ) w =z + Ll

C)

e) w = eZ h) w = sinhz

j) w = ezcosz 2. Evaluate each of the following limits: sinz + z a) limz+ni eC+2 C) limz,o

3. Differentiate each of the following complex functions:

4. Prove the rule (8.20) 5. Prove Theorem 1.

+

b) 1imz+o d) l i m z + ~ z

, , sin z

541

Chapter 8 Functions of a Complex Variable

Figure 8.3

') f

Complex line integral.

,

to bt

The complex integral ff ( z )d z is defined as a line integral, and its properties are closely related to those ,of the integral [ P d x Q d y (see Chapter 5). Let C be apath from A to B in the complex plane: x = x ( t ) , y = y ( t ) , a 5 t .5 b. We assume C to have a direction, usually that of increasing t . We subdivide the interval a 5 t 5 b into n parts by to = a , tl , . . . , t,, = b. We let z, = x ( t j ) i y ( t j ) and A j z = z , - z j - , , A j t = t j - t j - , . We choose an arbitrary value t j in the 5 t 5 ti and set zy = x(r;) i y ( t f ) . These quantities are all shown in interval Fig. 8.3. We then write

+

+

+

If we take real and imaginary parts in (8.21), we find

f ( 2 ) d z = lirn x ( u + i u ) ( A x + i A y ) C

that is,

The complex line integral is thus simply a combination of two real line integrals. Hence we can apply all the theory of real line integrals. In the following, each path

542

Advanced Calculus, Fifth Edition

8

is assumed to be piecewise smooth; that is, x ( t ) and y ( t ) are to be continuous with piecewise continuous derivatives. As in Section 5.2, we have: THEOREM 5 and

If f (z) is continuous in domain D, then the integral (8.21) exists,

We now write our path as z = z ( t ) as in Section 8.2. If we introduce the derivative

of z with respect to the real variable t and also use the theory of integrals of such functions (Section 8.2), we can write (8.23) more concisely:

EXAMPLE I

Let C be the path x = 2 t , y = 3t, 1 5

t

5 2. Let f ( 2 ) = z 2 . Then

EXAMPLE 2 Let C be the circular path x = cos t , y = sin t , 0 5 t 5 2n. This . d z l d t = ieir, can be written more concisely thus: z = eir,0 5 t 5 2 ~Since

Further properties of complex integrals follow from those of real integrals: THEOREM 6 Let f ( z )and g(z) be continuous in a domain D. Let C be a piecewise smooth path in D. Then

Chapter 8 Functions of a Complex Variable where C is composed of a path C1 from zoto zl and a path C2 from ZIto 22, and '9

where C' is obtained from C by reversing direction on C. Upper estimates for the absolute value of a complex integral are obtained by the following theorem.

THEOREM 7 Let f (z)be continuous on C, let If (z)l IM on C, and let

be the length of C. Then

Proof.

The line integral j

If

(z)l ds is defined as a limit:

where A,s is the length of the jth arc of C. Now

for I A ,z 1 represents the chord of the arc A ,s. Hence

by repeated application of the triangle inequality (8.1). Passing to the limit, we conclude that

Also, if If 1

_(

*

M = const,

C If(z;)l A j s 5

1

M A ~ S= M L .

Hence

C

Inequalities (8.25)follow from (8.26)and (8.27).

544

Advanced Calculus, Fifth Edition

PROBLEMS 1. Evaluate the following integrals:

a) $+'(x2 - i Y 2 )d z on the straight line from 0 to 1 b) 1; z d z on the curve y = sin x .

+i.

A' 9

c) on the linex = 1. 2. Write each of the following integrals in the form J u d x - v d y i J v d x u d y ; then show that each of the two real integrals i~ ~ndependentof path in the xy-plane. a) J ( z 1 ) d z b) J e Z d z C) J z 4 d z d) S s i n z d z 3. a) Evaluate

+

+

+

on the circle lzl = R. b) Show that

on every simple closed path not meeting or enclosing the origin. c) Show that

on every simple closed path not passing through the origin.

A function w = f (z), defined in a domain D, is said to be an analytic function in D if w has a continuous derivative in D. Almost the entire theory of functions of a complex variable is confined to the study of such functions. Furthermore, almost all functions used in the applications of mathematics to physical problems are analytic functions or are derived from such. It will be seen that possession of a continuous derivative implies possession of a continuous second derivative, third derivative, . . . , and, in fact, convergence of the Taylor series

in a neighborhood of each zo of D. Thus one could define an analytic function as one so representable by Taylor series, and this definition is often used. The two definitions are equivalent, for convergence of the Taylor series in a neighborhood of each zo implies continuity of the derivatives of all orders. Although it is possible to construct continuous functions of z that are not analytic (examples will be given), it is impossible to construct a function f ( z ) possessing a derivative, but not a continuous one, in D. In other words, if f ( z ) has a derivative in D, the derivative is necessarily continuous, so that f ( z ) is analytic. Therefore we could define an analytic function as one merely possessing a derivative in domain

Chapter 8 Functions of a Complex Variable

D, and this definition is also often used. For a proof that existence of the derivative implies its continuity, refer to Vol. I of the book by Knopp listed at the end of the chapter. THEOREM 8 If w = u + i v = f ( z )is analytic in D, then u and v have continuous first partial derivatives in D and satisfy the Cauchy-Riemann equations

in D. Furthermore,

Conversely, if u ( x , y ) and v ( x , y ) have continuous first partial derivatives in D and satisfy the Cauchy-Riemann equations (8.28),then w = u i v = f ( z ) is analytic in D.

+

Proof. Let zo be a fixed point of D and let

as in Fig. 8.4. We consider several equivalent formulations of the condition that f '(zo) exists. Throughout, 6 , e l , r 2 ,E ~ c4, denote functions of Az = Ax i A y , continuous and equal to zero at Az = 0. By Theorem 3 , existence of f ' ( 2 0 ) is equivalent to the statement

+

Aw = c . Az

+

6

. Az,

this is equivalent to

and also to

Figure 8.4 %

Complex derivative.

c = fl(zo),

c =a

+ ib;

(8.30)

545

546

Advanced Calculus, Fifth Edition where r l , €2, €3, €4 are real. For if (8.30') holds, then (8.30") holds with €1 =Re (r), r 2 = -1m (c), €3 = Im (r ), c4 = Re (c). Conversely, if (8.30") holds, then (8.30') holds with r = 0 for Az = 0 and

As Fig. 8.4 shows,

so that we deduce from (8.3 1) that r + 0 as Az + 0. Thus (8.30), (8.301),and (8.30") are all equivalent to existence of f '(zo) = c = a ib. By taking real and imaginary parts in (8.30") we obtain one more equivalent condition:

+

these equations state that u , v have differentials du = a dx -b dy , dv = b dx +a dy at (xo,yo), and hence at this point 5

Thus differentiability of f '(2) at any z is equivalent to differentiability of u , v along with validity of the Cauchy-Riemann equations. Furthermore, f '(z) and a u / a x , . . . are related by (8.29). By Theorem 1 these equations show that continuity off '(z) in D is equivalent to continuity of a u l a x , . . . . Thus the theorem is proved. The theorem provides a perfect test for analyticity: If f (z) is analytic, then the Cauchy-Riemannequations hold; if the equationshold (and the derivatives concerned are continuous), then f (z) is analytic. EXAMPLE 1

w =z2=x2-y2+i.2xy.Hereu=x2-

y2,v=2xy.Thus

and w is analytic for all z. EXAMPLE 2

w =

A - -A. Here x +y x +y

au -- y2 -x*

a~ --

a ~-. -

-2xy av - -(x2 y2)2 ax'

+ y*)2 a ~ ' ay + Hence w is analytic except for x2 + y2 = 0, that is, for z = 0. a~

EXAMPLE3

(x2

w=x-iy=Z.Hereu=x,v=-y,and

Thus w is not analytic in any domain.

Chapter 8 Functions of a Complex Variable

EXAMPLE 4

w = x~~~ + 2x2y2i.Here

The Cauchy-Riemann equations give 2xy2 = 4x2y,2x2y = -4xy2. These equations are satisfied only along the lines x = 0,y = 0. There is no domain in which the Cauchy-Riemann equations hold, hence no domain in which f (z) is analytic. One does not consider functions analytic only at certain points unless these points form a domain. The terms analytic at a point or analytic along a curve are used, apparently in contradiction to the remark just made. However, we say that f (z) is analytic at the point zo only if there is a domain containing zo within which f (z) is analytic. Similarly, f (z) is analytic along a curve C only if f (z) is analytic in a domain containing C. THEOREM 9 The sum, product, and quotient of analytic functions are analytic (provided that in the last case the denominator is not equal to zero at any point of the domain under consideration). All polynomials are analytic for all z. Every rational function is analytic in each domain containing no root of the denominator. An analytic function of an analytic function is analytic.

This follows from Theorem 4. We readily verify (Problem 1 below) that the Cauchy-Riemann equations are satisfied for u = Re (eZ),v = Im (eZ).Hence eZis analytic for all z. It then follows from Theorem 9 that sin z, cos z, sinh z, and cosh z are analytic for all z, while tan z, sec z, and csc z are analytic except for certain points (Problem 6 below). Furthermore, the usual formula for derivatives holds:

(Problem 3). Two basic theorems of more advanced theory are useful at this point. Proofs are given in Chapter IV of the book by Goursat listed at the end of the chapter.

THEOREM 10 Given a function f (x) of the real variable x , a < x I: b, there is at most one analytic function f (z) that reduces to f (x) when z is real. THEOREM 11 If f (z), g(z), . . . are functions that are all analytic in a domain D that includes part of the real axis, and f (z), g(z), . . . satisfy an algebraic identity when z is real, then these functions satisfy the same identity for all z in D. Theorem 10 implies that our definitions of eZ,sin z, . . . are the only ones that yield analytic functions and agree with the definitions for real variables. Because of Theorem 1 1, we can be sure that all familiar identities of trigonometry, namely,

547

548

Advanced Calculus, Fifth Edition continue to hold for complex z. A general algebraic identity is formed by replacing the variables wl , . . . , w, in an algebraic equation by functions f ~ ( z ).,. . , f,(z). Thus in the two examples given, one has W:

+ W; - 1 = 0

(w, = sinz, w2 = COSZ),

To prove identities such as ezl

. ez' = eZl+Z?,

(8.33)

it may be necessary to apply Theorem 11 several times. (See Problems 4 and 5 below.) It should be remarked that although eZis written as a power of e, it is best not to think of it as such. Thus ell2 has only one value, not two, as would a usual complex root. To avoid confusion with the general power function, to be defined later, we often write e = exp z and refer to e: as the exponential3~nctionof z. To obtain the real and imaginary parts of sin z, we use the identity sin (2 I

+ z2) = sin z

cos 22

+ COS 21 sin 22,

which holds, by the reasoning described above, for all complex z l and z2. Hence sin (x iy) = sin x cos iy cos x sin iy . Now from the definitions (Section 8. I),

+

+

sinhy = -i siniy, cosh y = cos iy.

1

I

I

Hence sinz = sinxcoshy +icosxsinhy.

(8.35)

Similarly, we prove, as in (8.13), cosz = cosxcoshy -isinxsinhy, sinhz = sinhxcosy +icoshxsiny,

I

coshz = coshx cos y + i sinhx sin y.

(8.36)

,

'It

.re:

Conformal mapping. A complex function w = f ( z ) can be con&dered as a mapping from the xy-plane to the uv-plane as in Section 2.7. In the case of an analytic function f (z) this mapping has a special property: It is a conformal mapping. By this we mean that two curves in the xy-plane, meeting at (xo,yo) at angle cr, correspond to two curves meeting at the corresponding point (uo, vo) at the same angle a! (in value and in sense-positive or negative). This means that a small triangle in the xy-plane corresponds to a similar small (curvilinear) triangle in the uv-plane. (The properties described fail at the exceptional points where f '(z) = 0.) Furthermore, every conformal mapping from the xy-plane to the uv-plane is given by an analytic function. For a discussion of conformal mapping and its applications, see Sections 8.20 to 8.27.

549

Chapter 8 Functions of a Complex Variable

PROBLEMS 1. Verify that the following are analytic functions of z: a) 2x3 - 3x2y - 6xy2 y3 i(x3 6x2y - 3xy2 - 2y3) b) w = e Z =eXcosy+iexsiny C) w = sinz = sinxcoshy + i cosx sinhy 2. Test each of the following for analyticity: a) x3 y3 i(3x2y 3xy2) b) sinxcosy +icosxsiny c) 3x 5y i(3y - 5x) 3. Prove the following properties directly from the definitions of the functions:

+ +

+ +

+

+

+ +

.

..

d a) $-ez = ez b, az sinz =cosz, z c o s z = -sinz c) sin (z + n) = -sin z d) sin (-z) = -sin z, cos (-z) = cos z 4. Prove the identity eZ1+Z2= eZl . eZ2by application of Theorem 11. [Hint: Let z2 = b, a fixed real number, and zl = z, a variable complex number. Then eL+b= eL- eb is an identity connecting analytic functions which is known to be true for z real. Hence it is true for all complex 2. Now proceed similarly with the identity eZ1+Z = eZ1. eZ.] '11 N 5. Prove the following identities by application of Theorem 11 (see Problem 4): ~7 a) cos (zl z2) = cos Z I cos z2 - sin zl sin z2 b) elz = cos z i sin z C) (eZ)"= enZ (n = 0, 1,2, . . .) 6. Determine where the following functions are analytic (see Problem 3 following Section 8.2): b) C O ~ Z= cos z c) tanh z = sinh z a) tanz = sin z

+

d)

~iy

i

+

, ,

e)

i ez- z ~

eL

f ) sinz+cosz

8.7 THEFUNCTIONS log z ,aZ,za ,sin-' z ,cos-' z The function w = log z is defined as the inverse of the exponential function z = ew. We write z = reie, in terms of polar coordinates r, 0 , and w = u + i v , so that reiQ

- e ~ +=i ~e u e i v ,

Accordingly,

>UK

w =,log2 = logr

+ i(8 + 2kn) = log lzl + i argz,

(8.37)

where log r is the real logarithm of r. Thus log z is a multiple-valued function of z, with infinitely many values except for z = 0. We can select one value of 0 for each z and obtain a single-valued function, log z = log r i8; however, 8 cannot be chosen to depend continuously on z for all z # 0, since 0 will increase by 217 each time one encircles the origin in the positive direction. If we concentrate on an appropriate portion of the z-plane, we can choose 8 to vary continuously within the domain. For example, the inequalities

+

550

Advanced Calculus, Fifth Edition

Figure 8.5

Domain for log z.

together describe a domain (Fig. 8.5) and also tell how to assign the values of 0 within the domain. With 0 so restricted, logr i0 then defines a branch of logz in the domain chosen; this particular branch is called the principal value of logz and is denoted by Log z. The points on the negative real axis are excluded from the domain, but we usually assign the values Log z = log 1x1 in on this line. Within the domain of Fig. 8.5, Log z is an analytic function of z (Problem 4 below). Other branches of log z are obtained by varying the choice of 0 or of the domain. For example, in the domain of Fig. 8.5 we might choose 0 so that n < 0 < 3n or so that -3n < 0 < -n. The inequalities 0 < 8 < 2n, n/2 < 0 < 5n/2,. . . also suggest other domains and choices of 8. We can verify that as long as 0 varies continuously in the domain, log z = log r + iQis analytic there. The most general domain possible here is an arbitrary simply connected domain not containing the origin. As a result of this discussion it appears that log z is formed of many branches, each analytic in some domain not containing the origin. The branches fit together in a simple way; in general, we can get from one branch to another by moving around the origin a sufficient number of times while varying the choice of log z continuously. We say that the branches form "analytic continuations" of each other. We can further verify that for each branch of log z the rule

+

+

remains valid. The familiar identities are also satisfied (Problems 4 and 5). Before turning to exponential functions we recall the rule of algebra for the nth roots of a complex number z = r(cos Q i sin 0):

+

z'ln = [r(cos0

+ i sin 0)]'/"

The rule follows from the Demoivre formula: [r(cos 0

+ i sin O)]" = rn(cosno + i sin no).

The general exponentialfunction a Zis defined, for a

= ,z

logs

- exp (z log a).

# 0,by the equation (8.40)

Chapter 8 Functions of a Complex Variable

+

Thus for z = 0,a0 = 1. Otherwise, log a = log la\ i arg a , and we obtain many values: a: = exp [z(log Jal i(a 2nn))], (n = 0, f1, f2, . . .), where a! denotes one choice of arg a . For example,

+

+

- e-(n/4)-2nn

(COSlog z/Z + i sin log

A).

If z is a positive integer rn, ai reduces to a m and has only one value. The same holds for z = -m, and we have

If z is a fraction p l q (in lowest terms), we find that a' has q distinct values, which are the 9th roots of U P . (See Eq. (8.39).) If a fixed choice of log a is made in (8.40), then a 3 s simply erZ,c = loga, and is hence an analytic function of z for all z. Each choice of log a determines such a function. If a and z are interchanged in (8.40), we obtain the general power function, Z~ = eo :%1

(8.42)

If an analytic branch of logz is chosen as above, then this function becomes an analytic function of an analytic function and is hence analytic in the domain chosen. In particular, the principal value of zu is defined as the analytic function z" = e" L"G, in terms of the principal value of log z. For example, if a = we have

i,

, = fl(z) becomes analytic in the as in Eq. (8.39) with k = 0. If Log z is used, then h domain of Fig. 8.5. A second analytic branch f2(z) in the same domain is obtained by requiring that n < 9 < 3 n . These are the only two analytic branches that can be obtained in this domain. It should be remarked that these two branches are related by the equation f2(z) = -fl(z). For f 2 is obtained from f l by increasing 9 by 27r, which replace e(l/3'e by 1/2)iO = -e(1/2)iB e(1/2)1'(8+2a)= The functions sin-' z and cos-' z are defined as the inverse of sin z and cos z. We then find 1 sin-' z = : log [ i z f J-

1, 1 z = log [z fi JS 1. 1

cos-'

1

The proofs are left to Problem 2 below. It can be shown that analytic branches of both these functions can be defined in each simply connected domain not

551

552

Advanced Calculus, Fifth Edition

4-

containing the points dZ1. For each z other than dZ 1, one has two chbices of and then an infinite sequence of choices of the logarithm, differing by multiples of 2ni.

PROBLEMS 1. Obtain all values of each of the following: a ) log2 e) (1

+ i)'"

c ) log(1 - i ) g) sin-' 1

b ) log i f ) iJZ

d) if h) cos-' 2

2. Prove the formula (8.43). [Hint: If w = sin-' z, then 2iz = elw - e-lW;multiply by ef and solve the resulting equation as a quadratic for e l W . ] 3. a ) Evaluate sin-' 0 , cos-' 0. b) Find all roots of sin z and cos z [compare part (a)]. 4. Show that each branch of log z is analytic in each domain in which 0 varies continuously and that ( d l d z )log z = 112. [Hint: Show from the equations x = r cos8, y = r sin0 that a0lax = - y / r 2 , a8/ay = x / r 2 . Show that the Cauchy-Riemann equations hold for u = log r, v = 8.1 5. Prove the following identities in the sense that for proper selection of values of the multiplevalued functions concerned the equation is correct for each allowed choice of the variables: a ) log(zl . z 2 ) = log z l log z 2 ( z i # 0, z 2 # 0 ) b) elogz = z ( Z # 0 ) C ) log eZ = z

+

d ) log z1z2 = z2 log zi ( z i # 0 )

6. For each of the following, determine all analytic branches of the multiple-valued function in the domain given:

a) logz, x c 0

b)

firx > O

7. Prove that for the analytic function za (principal value),

8. Plot the functions u = Re ( A ) and v = Im (A) as functions of x and y and show the two branches described in the text. 9. Let w = u + i v = f ( z ) , z = x + i y , be analytic in domain D. a ) Show that a(u, v ) / a ( x , y ) =' 1 f'(z)12. b) Apply the inverse mapping theorem of Section 2.12 to conclude that if f l ( z o ) # 0 , where zo is in D , and f ( z o ) = wo, then there is an inverse z = g ( w ) of f defined in a neighborhood Do of wo, mapping Do onto a neighborhood of zo. C) Show that g ( w ) is analytic in Do. [Hint: Find the corresponding Jacobian matrix and show that the Cauchy-Riemann equations in u , v are satisfied by x ( u , v ) = Re d w ) , y(u, v ) = Im g(w).l d ) Show that g l ( w ) = l l f ' ( z )at w = f ( z ) and that g l ( w ) # 0 in Do.

Chapter 8 Functions of a Complex Variable

Figure 8.6

Cauchy integral theorem.

All paths in the integrals concerned here, as elsewhere in the chapter, are assumed to be piecewise smooth. The following theorem is fundamental for the theory of analytic functions:

THEOREM 12 (Cauchy Integral Theorem) If f (2) is analytic in a simply connected domain D, then

on every simple closed path C in D (Fig. 8.6).

Proof.

We have, by (8.22), udx - v d y + i

i

vdx +udy.

The two real integrals are equal to zero (see Section 5.6) provided that u and v have continuous derivatives in D and

These are just the Cauchy-Riemann equations. Hence

C

This theorem can be stated in an equivalent form:

THEOREM 12' If f ( z ) is analytic in the simply connected domain D, then 1f (2) d z is independent of the path in D.

553

554

Advanced Calculus, Fifth Edition For independence of path and equaling zero on closed paths are equivalent properties of line integrals. If C is a path from z l to z2, we can now write

the integral being the same for all paths C from zl to z 2 .

+

THEOREM 13 Let f ( z ) = u i v be defined in domain D and let u and v have continuous partial derivatives in D. If

on every simple closed path C in D, then f ( z ) is analytic in D. Proof. The condition (8.44) implies that

udx -vdy =0,

' .

C

vdx+udy=O C

on all simple closed paths C; that is, the two real line integrals are independent of path in D. Therefore by Theorem I11 in Section 5.6,

since the Cauchy-Riemann equations hold, f is analytic. This theorem can be proved with the assumption that u and v have continuous derivatives in D replaced by the assumption that f is continuous in D; it is then known as Morera's theorem. For a proof, see Chapter 5 of Vol. I of the book by Knopp listed at the end of the chapter.

[r

THEOREM 14 If f ( z ) is analytic in D, then f l ( z ) d z = f(z)lii = f ( r 2 ) - ~ ( Z I )

on every path in D from z l to z2. In particular,

on every closed path in D. Proof.

By (8.29),

(8.45)

Chapter 8 Functions of a Complex Variable This rule is the basis for evaldation of simple integrals, just as in elementary calculus. Thus we have

In the first of these, any path can be used; in the second, any path not through the origin. THEOREM 15

I f f (z) is analytic in D and D is simply connected, then F(z) =

[:f (z) dz

(ZI

fixed in D)

is an indefinite integral of f (2); that is, F'(z) = f (2). Thus F(z) is itself analytic.

L:

Proof. Since f (z) is analytic in D and D is simply connected, f (z) dz is independent of path and defines a function F that depends only on the upper limit z. We have, further, F = U i V , where

+

+

and both integrals are independent of path. Hence d U = u dx - v d y , d V = v d x u d y . Thus U and V satisfy the Cauchy-Riemann equations, so that F = U i V is analytic and

+

Cauchy 's theorem for multiply connected d o k i n s . I f f (z) is analytic in a multiply connected domain D, then we cannot conclude that

on every simple closed path C in D. Thus if D is the doubly connected domain of Fig. 8.7 and C is the curve C 1 shown, then the integral around C need not be zero. However, by introducing cuts we can reason that

that is, the integral has the same value on all paths that go around the inner "hole" once in the positive direction. For a triply connected domain, as in Fig. 8.8, we obtain

555

r

556

Advanced Calculus, Fifth Edition

Figure 8.7

Cauchy theorem for doubly connected domain.

Figure 8.8

Cauchy theorem for triply connected domain.

t

the equation

This can be written in the form

Equation (8.49) states that the integral around the complete boundary of a certain region in D is equal to zero. More generally, we have the following theorem:

THEOREM 16 (Cauchy's Theorem for Multiply Connected Domains) Let f ( z ) be analytic in a domain D and let C , , . . . , C, be n simple closed curves in D that together form the boundary B of a region R contained in D. Then

1

~ ( Z M Z= 0,

B

where the direction of integration on B is such that the outer normal is 90" behind the tangent vector in the direction of integration.

Chapter 8 Functions of a Complex Variable

Now let D be a simply connected domain and let zo be a fixed point of D. If f ( z ) is analytic in D, the function f ( z ) / ( z - zo) will fail to be analytic at zo. Hence

will in general not be zero on a path C enclosing zo. However, as above, this integral will have the same value on all paths C about zo. To determine this value, we reason that if C is a very small circle of radius R about zo, then f ( z o ) has, by continuity, approximately the constant value f ( z o ) on the path. This suggests that

since we find

with the aid of the substitution: z - zo = ~ e " . The correctness of the conclusion reached is the content of the following fundamental result: , THEOREM 17 (Cauchy Integral Formula) Let f ( z ) be analytic in a domain D. Let C be a simple closed curve in D, within which f ( z ) is analytic and let zo be inside C. Then

Proof. The domain D is not required to be simply connected, but since f is analytic within C, the theorem concerns only a simply connected part of D, as shown in Fig. 8.9. We reason as above to conclude that

Figure 8.9

Cauchy integral formula.

557

558

Advanced Calculus, Fifth Edition It remains to show that the integral on the right iiTindeed f(zoJ'. 2ni. Now, since f (20) = const,

where we integrate always on the circle lz - zol = R. Hence on the same path,

Now lz - zol = R on the path, and since f ( z ) is continuous at zo, 1 f ( z ) - f (zo)l < E for R < 6, for each preassigned E > 0 . Hence by Theorem 7,

Thus the absolute value of the integral can be made as small as desired by choosing R sufficiently small. But the integral has the same value for all choices of R. This is possible only if the integral is zero for all R. Hence the left-hand side of (8.51) is zero, and (8.50) follows. I I??-< The integral formula (8.50) is remarkable in that it expresses the values of the function f ( z ) at points zo inside the curve C in terms of the values along C alone. , (8.50) reduces to the following: If C is taken as a circle z = zo ~ e " then

+

Thus the value of an analytic function at the center of a circle equals the average (arithmetic mean) of the values on the circumference. Just as with the Cauchy integral theorem, the Cauchy integral formula can be extended to multiply connected domains. Under the hypotheses of Theorem 16,

where zo is any point inside the region R bounded by C, (the outer boundary), C2, . . . , C,. The proof is left as an exercise (Problem 6).

PROBLEMS 1. Evaluate the following integrals: a) $z2 sin z dz on the ellipse x 2 2 y 2 = 1

+

b)

$-&

dz on the circle lz - 21 = 1

,I.

Chapter 8

559

Functions of a Complex Variable

' c) J,2f zeZ dz on the line segment joining the endpoints

d) j " ; ~f ~dz on the parabola zy2 = r

+1

2. a) Evaluate Ji,(dz/z) on the path z = ei', -n/2 5 t 5 n/2, with the aid of the relation (log z)' = l/z, for an appropriate branch of log z. b) Evaluate A-'(dzlz) on the path z = elt, n / 2 5 t 5 3x12, as in part (a). C) Why does the relation (log 2)' = l/z not imply that the sum of the two integrals of parts (a) and (b) is zero? 3. A certain function f (z) is known to be analytic except for z = 1, z = 2, z = 3, and it is known that

CA

where Ckis a circle of radius

with center at z = k. Evaluate

on each of the following paths: 4 ,, b) lzl = 2.5 c) Iz 2.51 = 1 a) lzl = 4 4. A certain function f ( 2 ) is analytic except for z = 0, and it is known that lim zf (z) = 0.

+

z+m

Show that

f (2) dz = 0

t

on every simple closed path not passing through the origin. [Hint: Show that the value of the integral on a path lzl = R can be made as small as desired by making R sufficiently large.] 5. Evaluate each of the following with the aid of the Cauchy integral formula: a) e)

f& $*

f f i dz on lzl = 1 d) f* dz on lzl = 2 z +1

d r on lzl = 5 dz on

b)

IZI = 2

[Hint for (c) and (d): Expand the rational function in partial fractions.] 6. Prove (8.53) under the hypotheses stated. 7. Prove that if f (z) is analytic in domain D and f '(2) = 0,then f (z) = constant. [Hint: Apply Theorem 14.1

We now proceed to enlarge the class of specific analytic functions still further by showing that every power series

converging for some values of

z

other than

z

= zo represents an analytic function.

1

560

Advanced Calculus, Fifth Edition For the theory of series of complex numbers, see Section 6.19. The following fundamental theorem for complex power series is proved just as for real series (Theorem 35 in Section 6.15). cn(z o - zo)" has a radius of convergence THEOREM 18 Every power series ESPE r* such that the series converges absolutely when (z - zo( < r * and diverges when lz -zoI > r*. The series converges uniformly for lz - zoI 5 rl, provided that rl < r*. The number r* can be zero, in which case the series converges only for z = zo, a positive number, or oo, in which case the series converges for all z. The number r* can be evaluated as follows:

1% 1,

r* = lim "-+O0Cn+l

if the limit exists. I

.I

1

-+

(8.54)

r* = lim -2.- if the limit exists, n+w

m'

and in any case by the formula

As for real variables, no general statement can be made about convergence on the boundary of the domain of convergence. This boundary (when r* # 0,r* # oo) is a circle ) z - zol = r*, termed the circle of convergence (Fig. 8.10). The series may converge at some points, all points, or no points of this circle.

EXAMPLE 1

C z , ( z n / n 2 ) .The first formula (8.54) gives

The series converges absolutely on the circle of convergence, for when series of absolute value is the convergent series C ( l / n 2 ) .

Convergence

Figure 8.10

lzl = 1, the

D~vergence

Circle of convergence of a power series. P C

J

561

Chapter 8 Functions of a Complex Variable

CEO

EXAMPLE 2 zn. This complex geometric series converges for J z J< 1, as (8.54) shows. We have further

as for real variables. On the circle of convergence the series diverges everywhere, since the nth term fails to converge to zero. The following theorems are proved as for real variables. THEOREM 19 A power series with nonzero convergence radius represents a continuous function within the circle of convergence.

THEOREM 20 A power series can be integrated term by term within the circle of convergence; that is, if r* # 0 and ( l z - zol < r * ) , then, for every path C inside the circle of convergence,

or in terms of indefinite integrals,

+ const J f ( z ) d z = ccn(z n +zo)"+l l do

-

(Iz - 201 < r * ) .

n=O

I

THEOREM 21 and

then

A power series can be differentiated term by term; that is, if r* # 0

>*

562

Advanced Calculus, Fifth Edition Hence every power series with nonzero convergence radius defines an analytic function f (z) within the circle of convergence, and the power series is the Taylor series of f (2):

THEOREM 22 If two power series x r = o cn(z - zo)", Cn(z - 20)" have nonzero convergence radii and have equal sums wherever both series converge, then the series are identical; that is,

,

In Section 8.10 it was shown that every power series with nonzero convergence radius represents an analytic function. We now proceed to show that all analytic functions are obtainable in this way. If a function f (z) is analytic in a domain D of general shape, we cannot expect to represent f (z) by one power series, for the power series converges only in a circular domain. However, we can show that for each circular domain Do in D there is a power series converging in Do whose sum is f (2). Thus several (perhaps infinitely many) power series are needed to represent f (z) throughout all of D.

THEOREM 23 Let f (z) be analytic in the domain D. Let z0 be in D and let R be the radius of the largest circle with center at zo and having its interior in D. Then there is a power series

n=O that converges to f (z) for 1 z - zo I < R . Furthermore,

where C is a simple closed path in D enclosing zo and within which f (z) is analytic. Proof. For simplicity we take zo = 0. The general case can then be obtained by the substitution z' = z - zo. Let the circle lzl = R be the largest circle with center at zo and having its interior within D ; the radius R is then positive or +oo (in which case D is the whole z-plane). Let zl be a point within this circle, so that lzl I < R. Choose RZ so that lz, 1 < R2 < R (see Fig. 8.1 1). Then f (z) is analytic in a domain including the circle CZ: = R2 plus interior. Hence by the Cauchy integral formula,

Chapter 8 Functions of a Complex Variable

Figure 8.11

Now the factor

Taylor series of an analytic function.

l / ( z - z l ) can be expanded in a geometric series:

1 -2-21

-

1 z(1-y)

='(1+41+...+-+...

z

Z

2'; zn

The series can be considered as a power series in powers of l / z , for fixed z,. It converges for lzl /zl < 1 and converges uniformly for lzl /zl i Izl I/ Rz < 1. If we multiply by f ( z ) , we find

since f ( z ) is continuous for lzl = R2, the series remains uniformly convergent on C2. Hence we can integrate term by term on CZ:

The left-hand side is precisely f (zl), by the integral formula. Hence

The path C2 can be replaced by any path C as described in the theorem, since f (z)/zn+' is analytic in D except for z = zo = 0. By Theorem 2 1 the series obtained is the Taylor series of f , so that

The theorem is now completely proved. The consequences of this theorem are far-reaching. First of all, not only does it guarantee that every analytic function is representable by power series, but it also

563

564

Advanced Calculus, Fifth Edition ensures that the Taylor series converges to the function within each circular domain within the domain in which the function is given. Thus without further analysis we at once conclude that

for all z. A variety of other familiar expansions can be obtained in the same way. It should be recalled that a function f (z) is defined to be analytic in a domain D if f (z) has a continuous derivative f r ( z ) in D (Section 8.6). By Theorem 23, f (z) must have derivatives of all orders at every point of D. In particular, the derivative of an analytic function is itself analytic:

THEOREM 24 I f f (2) is analytic in domain D, then f '(z), f " ( z ) , . . . , f ("'(z), . . . exist and are analytic in D. Furthermore, for each n,

where C is any simple closed path in D enclosing zo and within which f (z) is analytic. Equation (8.57) is a restatement of (8.56). Circle of convergence of the Taylor series. Theorem 23 guarantees convergence of the Taylor series o f f (z) about each zo in D in the largest circular domain )z- zol < R in D, as shown in Fig. 8.1 1. However, this does not mean that R is the radius of convergence r* of the series, for r * can be larger than R, as suggested in Fig. 8.12. When this happens, the function f (z) can be prolonged into a larger domain, while retaining analyticity. For example, if f (z) = Log z (0 < 8 < n) is expanded in a

Figure 8.12

Analytic continuation.

Chapter 8 Functions of a Complex Variable

+

Taylor series about the point z = -1 i , the series has convergence radius &, whereas R = 1 [Problem 4(c)]. The process of prolonging the function suggested here is called analytic continuation. Harmonic functions.

Let w = f ( z ) be analytic in D and let

By Theorem 8 (Section 8.6), f '(2)= u, +i v, = vy-iu, = . . . , and by Theorem 24, f '(2)is analytic in D, so that Thus u,,, u,)., v,,, . . . are also continuous in D. In general, Theorem 24 shows that u and v have partial derivatives of all orders in D. Furthermore, by taking real and imaginary parts in the Taylor series of f ( z ) , as given in Theorem 23, we obtain convergent power series for u ( x , y ) and v ( x , y). For example, if f (z) = C c,zn for 121 < R, with c, = a, ib,, then

+

The same series can be written in terms of polar coordinates r, 8 by writing P, exp (i y,) (p, 2 O), zn = rn exp ( i d ) ,so that c,zn = p,rn exp [i(y,

c,

=

+ n8)I.

Hence we obtain for 0 5 r < R , 8 unrestricted, U(X,

y) = Re

(x

00

rn(ancos n0

c,zn) =

+ fin sin no),

n=O

with a , = p, cos y, , 3/, = -p, sin y, . (For fixed r , the series is a Fourier series in 8.) Also, by the Cauchy-Riemann equations, u, = v,, u , = -v,, so that

that is, U ( X , y ) is harmonic in D. Similarly v ( x ,y ) is harmonic in D. Conversely, it can be shown that, in a simply connected domain D, each harmonic function # ( x , y ) can be interpreted as the real or imaginary part of an analytic function in D. (See Problem 7.)

PROBLEMS 1. Determine the radius of convergence of each of the following series: 00 z18 (z C) CZon!zn d) CEO a) Cn=o b) CEO 2. Given the series r F , ( z n / n ) show , that a) the series has radius of convergence 1; b) the series diverges for z = 1; c) the series converges for z = i and for z = - 1 . It can be shown that the series converges for J z J= 1, except for z = 1.

5

565

Advanced Calculus, Fifth Edition 3. By means of (8.57), evaluate each of the following:

4. Expand in a Taylor series about the point indicated, and determine the radius of convergence r* and the radius R of the largest circle within which the series converges to the function: b ) l / ( z - 1)aboutz = 2

a ) sin z about z = 0

c) Logz(O
5. (Cauchy's inequalities) Let f (z) be analytic in a domain including the circle C: (z - zol = R and interior and let If (z)l 5 M = const on C. Prove that Mn! ~ f ' ~ ' ( z o )s l

R,

[Hint: Apply (8.57).] 6. A function f ( z ) that is analytic in the whole z-plane is termed an entire function or an integral function. Examples are polynomials, e Z ,sin z , cos z. Prove Liouville's theorem: If f ( z ) is an entire function and 1 f ( z ) J5 M for all z, where M is constant, then f (z) reduces to a constant. [Hint: Take n = 1 in the Cauchy inequalities of Problem 5 to show that f ' ( z o )= 0 for every z0.1 7. Let @ ( x ,y) be harmonic in the simply connected domain D. Let

where ( x o , yo) is a fixed point of D and the line integral is taken on a path in D. a) Show that the line integral is independent of path, so that $(x, y ) is well defined. b ) Show that 4x = $,, 4, = -Ilr,, so that f ( z ) = 4 + i $ is analytic in D.

8.12 POWERSERIESIN POSITIVE AND NEGATIVE

POWERS LAURENT EXPANSION

,

We have shown that every power series C a n ( z - zo)" with nonzero convergence radius represents an analytic function and that every analytic function can be built up out of such series. It thus appears unnecessary to seek other explicit expressions for analytic functions. However, the power series represent functions only in circular domains and are hence awkward for representing a function in a more complicated type of domain. It is therefore worthwhile to consider other types of representations. A series of form

will also represent an analytic function in a domain in which the series converges. For the substitution z1 = 1/ ( z - z o ) reduces the series to an ordinary power series,

Chapter 8 Functions of a Complex Variable If this series converges for lzl I c r;, then its sum is an analytic function F(z1): hence the series (8.58) converges for

to the analytic function g(z) = F ( l / ( z - 20)). The value zl = 0 corresponds to z = oo,in a limiting sense; accordingly, we can also say that g(z) is analytic at oo and g(oo) = 0. This will be justified more fully in Section 8.14. The domain of convergence of the series (8.58) is the region (8.59), which is the exterior of a circle. It can happen that r; = oo,in which case the series converges for all z except zo; if r; = 0, the series diverges for all z (except z = oo,as above). If we add to a series (8.58) a usual power series

2

an(z - ZO)"= a0

n=O

converging for

(z - zol

+ al(z - ZO)+ . . . , .

< r;, we obtain a Sum

If r,* < r;, the sum converges and represents an analytic function f (z) in the annular domain: r; < lz - zol < r;, for each series has an analytic sum in this domain, so that the sum of the two series is analytic there. We can write this sum in the more compact form (after some relabeling)

though this should be interpreted as the sum of two series, as in (8.60). In this way we build up a new class of analytic functions, each defined in a ring-shaped domain. Every function analytic in such a domain can be obtained in this way: THEOREM 25

(Laurent's Theorem) Let f (z) be analytic in the ring: RI <

- zol <

R2.

Then

where

and C is any simple closed curve separating lzl = converges uniformly for RI < k~ 5 lz - zol 5 k2

RIfrom lzl = R2. The series R2.

568

Advanced Calculus, Fifth Edition

Figure 8.13

Laurent's theorem.

Proof. For simplicity we take zo = 0.Let z l be any point of the ring and choose r l , r2 SO that RI < rl c ( z l1 < r2 < R2, as in Fig. 8.13. We then apply the Cauchy integral formula in general form [Eq. (8.53)] to the region bounded by C I :lzl = rl and C2: Iz I = r2. Hence

The first term can be replaced by a power series

as in the proof of Theorem 23 (Section 8.11). For the second term, the series expansion

valid for J z l( > lzl = r l , leads similarly to the series

Hence

c 00

f(z1) =

n=-M

an.z;,

1 a, = 2ni

f

f ( z ) dz; znfl

CI

the path C2 or C 1can be replaced by any path C separating lz 1 = R, from lz 1 = R2, since the function integrated is analytic throughout the annulus. The uniform convergence follows as for ordinary power series (Theorem 18). The theorem is now established. Laurent's theorem continues to hold when R1 = 0 or R2 = ao or both. In the case R1 = 0 the Laurent expansion represents a function f ( z ) analytic in a deleted

Chapter 8 Functions of a Complex Variable neighborhood of zo, that is, in the circular domain lz - zol < R2 minus its center zo. If R2 = oo,we can say similarly that the series represents f (z) in a deleted neighborhood of z = oo.

8.13 ISOLATED SINGULARITIES OF AN ANALYTIC FUNCTION ZEROSAND POLES

,

Let f (z) be defined and analytic in domain D. We say that f (z) has an isolated singularity at the point zo i f f (z) is analytic throughout a neighborhood of zo except at zo itself; that is, to use the term mentioned at the end of the preceding section, f (z) is analytic in a deleted neighborhood of zo but not at zo. The point zo is then a boundary point of D and would be called an isolated boundarypoint (see Fig. 8.14). A deleted neighborhood 0 < lz - zol < R2 forms a special case of the annular domain for which Laurent's theorem is applicable. Hence in this deleted neighborhood, f (z) has a representation as a Laurent series:

The form of this series leads to a classification of isolated singularities into three fundamental types: Case I. No terms in negative powers of z - zo appeal: In this case the series is a Taylor series and represents a function analytic in a neighborhood of zo. Thus the singularity can be removed by setting f (zo) = ao. We call this a removable singularity of f (2). It is illustrated by

z sin-

z

- I - - +z2- - . .z4. 3!

5!

at z = 0. In practice we automatically remove the singularity by defining the function properly. 1

Figure 8.14

Isolated singularity.

569

Advanced Calculus, Fifth Edition Case II. A finite number of negative powers of z - zo appeal: Thus we have

Here f

( z ) is said to have a pole of order N at zo.'we can write f

(2) =

1 ( Z - zoIN

g(z) = a-N

g(z),

+ a - N + ~ (-z Z O ) + . . . ,

.t

(8.64)

so that g ( z ) is analytic for lz - zol < R2 and g(zo) # 0. Conversely, every function f ( z )representable in the form (8.64) has a pole of order N at zo. Poles are illustrated by rational functions of z , such as

which has poles of order 1 at fi and of order 3 at z = 1. The rational function

is called the principal part of f

( z ) at the pole zo. Thus f ( z ) - p ( z ) is analytic at zo.

EXAMPLE 1

f

(2)

ez cos z = - at z = 0.

z3

To obtain the Laurent series, we expand the numerator in a Taylor series:

Hence eZcosz --

A

z

-1+ - -1 - +1 z3 z2 3

....

Here the first two terms form the principal part; the pole is of order 3. EXAMPLE 2

We expand z/(z3

+ 2) in a Taylor series about z = - 1:

z

(Z

+ 1)2(z3 + 2)

---

4 +-(Z + +1 -1

IS+...

.

The first two terms form the principal part; the pole is of order 2.

Chapter 8 Functions of a Complex Variable Case Ill. Injinitely many negative powers of z - zo appeal: In this case, f ( z ) is said to have an essential singularity at zo. This is illustrated by the function f ( z ) = e'lZ

which has an essential singularity at z = 0. In Case I, f ( z ) has a finite limit at zo, and accordingly, If (z)l is bounded near 20; that is, there is a real constant M such that ( f ( z ) ( < M for z sufficiently close to 20.

In Case 11, lirn,,,,, f ( z ) = oo,and it is customary to assign the value oo (complex) to f ( z ) at a pole. At an essential singularity, f ( z ) has a very complicated discontinuity. In fact, for every complex number c we can find a sequence zn converging to zo such that limn,, f (2,) = c (see Problem 8 following Section 8.14). It follows from this that if I f (z)l is bounded near zo, then zo must be a removable singularity, and if lim f ( z ) = oo at 20, then zo must be a pole. If f ( z ) is analytic at a point zo, and f ( z o ) = 0, then zo is termed a root or zero of f ( z ) . Thus the zeros of sin z are the numbers nrr (n = 0, f 1, f2, . . .). The Taylor series about zo has the form

where N 2 1 and aN # 0, or else f ( z ) E 0 in a neighborhood of zo. It will be seen that the latter case can occur only if f ( z ) = 0 throughout the domain in which it is given. If now f ( z ) is not identically zero, then

We say that f ( z ) has a zero of order N or multiplicity N at zo. For example, 1 - cos z has a zero of order 2 at z = 0,since

I f f ( z ) has a zero of order N at zo, then F ( z ) = l / f ( z ) has a pole of order N at zo, and the converse is also true. For if f has a zero of order N, then

+

as above, with +(zo) 0. It follows from continuity that +(z) # 0 in a sufficiently small neighborhood of 20. Hence g ( z ) = 1 /+(z) is analytic in the neighborhood, and g ( z o ) # 0.Now in this neighborhood, except for 20,

so that F has a pole at zo. The converse is proved in the same way.

571

4 572

Advanced Calculus, Fifth Edition It remains to consider the case when f covered by the following theorem.

= 0 in a neighborGod of zo. This is

THEOREM 26 The zeros of an analytic function are isolated, unless the function is identically zero; that is, i f f ( z ) is analytic in domain D and f ( z ) is not identically zero, then for each zero zo of f ( z ) there is a deleted neighborhood of zo in which f ( z ) # 0.

The proof is given in Section 3.2 of the book by Ahlfors listed at the end of this chapter.

The complex number oo has been introduced several times in connection with limiting processes, for example, in the discussion of poles in the preceding section. In each case, oo has appeared in a natural way as the limiting position of a point receding indefinitely from the origin. We can incorporate this number into the complex number system with special algebraic rules:

+

Expressions such as oo oo, oo - oo, and oo/oo are not defined. A function f ( z ) is said to be analytic in a deleted neighborhood of oo if f ( z ) is analytic for lzl > R1 for some R 1 .In this case the Laurent expansion with R2 = oo and zo = 0 is available, and we have

If there are no positive powers of z here, f ( z )is said to have a removable singularity at oo, and we make f analytic at oo by defining f (oo) = ao: a-1 a-n f(z)=ao+-+...+-+... ( 1 ~ 1> R I ) . (8.68) z zn This is clearly equivalent to the statement that if we set zl = l / z , then f ( z )becomes a function of z l with removable singularity at zl = 0. If a finite number of positive powers occurs, we have, with N 2 1,

where @(z) is analytic at oo and @(GO) = a~ # 0. In this case, f ( z ) is said to have a pole of order N at oo. The same holds for f (1 / z l ) at z = 0. Furthermore, lim f ( z ) = ao.

z+

a0

(8.70)

573

Chapter 8 Functions of a Complex Variable If infinitely many positive powers appear, f (z) is said to have an essential singularity at z = oo. If f (z) is analytic at m as in (8.68) and f (oo) = a0 = 0, then f (z) is said to have a zero at z = m. If f is not identically zero, then necessarily some a-N # 0, and

Thus g(z) is analytic at oo, and g(oo) = a-N # 0. We say that f (z) has a zero of order (or multiplicity) N at oo.We can show that if f (2) has a zero of order N at oo, then 1 / f (z) has a pole of order N at oo, and the converse is also true. The significance of the complex number oo can be shown geometrically by the device of stereographic projection, that is, a projection of the plane onto a sphere tangent to the plane at z = 0, as shown in Fig. 8.15. The sphere is given inxyt-space by the equation

'

so that the radius is i.The letter N denotes the "north pole" of the sphere, the point (0,0, 1). If N is joined to an arbitrary point z in the xy-plane, the line segment Nz will meet the sphere at one other point P, which is the projection of z on the sphere. For example, the paints of the circle lzl = 1 project on the "equator" of the sphere, that is, the great circle t = As z recedes to infinite distance from the origin, P approaches N as limiting position. Thus N corresponds to the complex number oo. We refer to the z-plane plus the number oo as the extended z-plane. To emphasize that oo is not included, we refer to thejnite z-plane.

i.

Figure 8.15

Stereographic projection.

574

Advanced Calculus, Fifth Edition

PROBLEMS

,

:

.

." -

r,r

- I

1. For each of the following, expand in a Laurent series at the isolated singularity given and state the type of singularity: a) atz=0 b, z2(z - 3) a t z = 0 C) atz = 0 d) cscz a t z = O [Hint for (d): Write

1 1 a-1 + a o + . . . -z = -= sin z z - (z3/3 !) - . . z and determine the coefficients a _ , , ao, a1 , . . . so that CSC

+

2. For each of the following, find the principal part at the pole given: b) z 2 - 2 (Z = 0) a) z 2 + ? + l (Z =0) ez sin z

c) (z = 1) d, z2(z"z+ 1) ( 2 = 0) 3. For each of the following, expand in a Laurent series at z = oo and state the type of singularity: 1 1 1 a) 1 - z = -tl-(T/z) b) s 2 4. Let f (z) be a rational function in lowest terms:

The degree d of f (z) is defined to be the larger of rn and n. Assuming the fundamental theorem of algebra, show that f (z) has precisely d zeros and d poles in the extended z-plane, a pole or zero of order N being counted as N poles or zeros. 5. For each of the following, locate all zeros and poles in the extended plane (compare Problem 4): C) z ' + 3 z 2 + 3 z + 1 a) z 6. Let A(z) and B(z) be analytic at z = zo; let A(zo) # 0 and let B(z) have a zero of N at zo, so that

5

has a pole of order N at zo. Show that the principal part of f (z) at zo is

and obtain the next term explicitly. [Hint: Set

Multiply across and solve for C-N,C-N+], . . . .] 7. Prove Riemann's theorem: If If (z)l is bounded in a deleted neighborhood'of an isolated singularity 20, then zo is a removable singularity of f ( z ) . [Hint: Proceed as in Problem 6 following Section 8.11 with the aid of (8.62).] 8. Prove the Theorem of Weierstrass and Casorati: If zo is an essential singularity of f (z), c is an arbitrary complex number, and 6 > 0, then If (z) - cl < 6 for some z, in every

Chapter 8 Functions of a Complex Variable neighborhood of zo. [Hint: If the property fails, then I/[ f (z) - c] is analytic and bounded in absolute value in a deleted neighborhood of zo. Now apply Problem 7 and conclude that f ( z ) has a pole or removable singularity at zo.]

Let f (z) be analytic throughout a domain D except for an isolated singularity at a certain point zo of D. The integral

will not in general be zero on a simple closed path in D. However, the integral will have the same value on all curves C which enclose zo and no other singularity of f. This value, divided by 27ri, is known as the residue of f (z) at zo and is denoted by Res [f(z), zol. Thus F

1

Res [f (z), zol =

&f f (z) dz. -

where the integral is taken over any path C within which f (z) is analytic except at zo (Fig. 8.16). THEOREM 27

The residue of f (z) at zo is given by the equation Res C f (z),

(8.74)

zol = a-1,

where

is the Laurent expansion of f (z) at zo.

Proof.

By(8.62),

where C is chosen as in the definition of residue. Hence (8.74) follows t o

A YB.:'

Figure 8.16

Residue.

-,

9,

..

575

& 576

Advanced Calculus, Fifth Edition

Figure 8.17

Cauchy residue theorem.

*I

'

-

If C is a simple closed path in D, within which f (z) is analytic except for isolated singularities at z,, . . . , zk, then by Theorem 16,

ff(z)dz f f ( z ) d z + . . . + =

CI

C

f

f(z.)dz,

CI

ie

where C , encloses only the singularity at zl , C2 encloses only z2, . .. , as in Fig. 8.17. We thus obtain the following basic theorem:

THEOREM 28 (Cauchy's Residue Theorem) If f (z) is analytic in a domain D and C is a simple closed curve in D within which f (z) is analytic except for isolated singularities at 2 , . . . . , zk, then

ff (z)dz = 2irilRes f (2). [

211

+ . .. + Res L f (z), zkl).

(8.76)

C

This theorem permits rapid evaluation of integrals on closed paths, whenever it is possible to compute the coefficient a _ , of the Laurent expansion at each singularity inside the path. Various techniques for obtaining the Laurent expansion are illustrated in the problems preceding this section. However, if we wish only the term in (z -zo)-' of the expansion, various simplifications are possible. We give several rules here:

RULE I

At a simple pole 7.0 (that is, a pole of first order), Res [f (z), 201 = lim (z - 20)f (2). 2-*ZII

RULE I1

At a pole zo of order N (N = 2 , 3 , . . .),

where g(z) = (z - 2 0 )f~(2).

Chapter 8 Functions of a Complex Variable

RULE I11 If A(z) and B ( z ) are analytic in a neighborhood of zo, A(zo) # 0, and B(z) has a zero at zo of order 1, then

t:
has a pole of first order at zo, and

RULE IV If A(z) and B(z) are as in Rule 111, but B(z) has a zero of second order at zo, so that f ( z ) has a pole of second order at zo, then 6A' B " - 2 A B "' Res [f( z ) , zol = 3~,,2 (8.77) where A and the derivatives A', B", B"' are evaluated at zo.

Proofs of rules. Let f ( 2 ) have a pole of order N : 1 [a-N a-N+,(z- 20) . " I = f (z)=

-~

(Z

where

'"'

0

)

+

+ ~

1 (Z - ~

g(z)7 0

)

~

>

i

;'10%

g(z) = ( z - z

O

)

( ~z ) ,~

g(z0) = a-N

and g is analytic at zo. The coefficient of ( z - zo)-' in the Laurent series for f ( z ) is the coefficient of ( z in the Taylor series for g(z). This coefficient, which is the residue sought, is g'N-"(zo) g'N-"(z) = lim z+zo ( N - I)! ' (N - 1 For N = 1 this gives Rule I; for N = 2 or higher we obtain Rule 11. Rules I11 and IV follow from the identity of Problem 6 following Section 8.14:

For a first-order pole, N = 1, and the residue is a0 - ---A(z0) bi Bf(zo)' For a second-order pole, N = 2, and the residue is [(albz- aob3)/bi].Since

this reduces to the expression (8.77).

577

I

1

578

4

Advanced Calculus, Fifth Edition

EXAMPLE 1

. t.,

f

g

d

4

.

z

.

;

v,r

F-Y-

i

-

--*

z = 2ni(Res[f(z). 11 +Res[f(z), -11).

1z1=2

Since f (z) has first-order poles at h1,we find, by Rule I,

-

zeZ = e Res [f(z), 11 = lim (z - 1) . zez = lim z-+ 1 22-1 z-+lz+l 2' TY

Res[f(z),-l]=

l i m ( z + l ) . - = zeZ z2- 1

z-+-1

lim -zeZ - - --e-1 z+-1z- 1 -2

'

Accordingly,

Rule 111 could also have been used: Res[f(z),I]=el - 2e, Res[f(z),-I]=22 z=1 This is simpler than Rule I, since the expression A(z)/B'(z), once computed, serves for all poles of the prescribed type. EXAMPLE 2

r t

~

All poles are of first order. Rule 111 gives A(z)/B1(z) = z/(4z3) = 1/(4z2) as the expression for the residue at any one of the four points. Moreover, z4 = 1 at each pole, so that

Hence

EXAMPLE 3

$

121=2

ez z(z-

dz = 2ni(Res [f(z), 01

+ Res [f (z), I]}.

At the first-order pole z =0, application of Rule I gives the residue 1. At the second-order pole z = 1, Rule I1 gives

Chapter 8 Functions of a Complex Variable Rule IV could also be used, with A = eZ,B = z3 - 2z2

Accordingly,

LU

--

+ z:

- "

8.16 RESIDUEAT INFINITY Let f (z) be analytic for Izl > R. The residue offz) at oo is defined as follows:

where the integral is taken in the negative direction on a simple closed path C, in the domain of analyticity off (z), and outside of which f (z) has no singularity other than oo.This is suggested in Fig. 8.18. Theorem 27 has an immediate extension to this case:

THEOREM 29

The residue of f ( z ) at oo is given by the equation Res [f (z),ool = -a-I,

where

is the coefficient of z-' in the Laurent expansion of f (z) at CQ: a-n f ( z ) = ...+--+... + -a-1 +ao+alz+.-..

zn

z

(8.78)

(8.79)

The proof is the same as for Theorem 27. It should be stressed that the presence of a nonzero residue at oo is not related to presence of a pole or essential singularity at oo.That is, f (2) can have a nonzero residue whether or not there is a pole or essential singularity, for the pole or essential singularity at oo is due to the positive powers of z, not to negative powers (Section 8.14). Thus the function ellz = 1 z-' (2!z2)-I . is analytic at CQ but has the residue - 1 there.

+

+

+

Figure 8.18

Residue at infinity.

579

dl4Eefi Advanced Calculus, Fifth Edition

Figure 8.19

Residue theorem for exterior.

I )

Cauchy's residue theorem has also an extension to include oo:

THEOREM 30 Let f (z) be analytic in a domarn D that'includes a deleted neighborhood of oo. Let C be a simple closed path in D outside of which f (2) is analytic except for isolated singularities at zl , . . . , z k . Then

f f (z)dZ

= Zxi(Res [f(z), zll

+ . . . + Res L f (z), z h + Res [f (z), m ] )

(8.80)

C

The proof, which is like that of Theorem 28, is left as an exercise (Problem 4 following Section 8.18). It is to be emphasized that the integral on C is taken in the negative direction (see Fig. 8.19) and that the residue a t co must be included on the right. For a particular integral

on a simple closed path C we have now two modes of evaluation: The integral equals 2rri times the sum of the residues inside the path (provided that there are only a finite number of singularities there), and it also equals minus 2rri times the sum of the residues outside the path plus that at 00 (provided there are only a finite number of singularities in the exterior domain). We can evaluate the integral both ways to check results. The principle involved here is summarized in the following theorem: THEOREM 31 If f (z) is analytic in the extended z-plane except for a finite number of singularities, then the sum of all residues of f (z) (including oo) is zero. a

To evaluate the residues at oo, we can formulate a set of rules like the preceding ones. However, the following two rules are adequate for most purposes:

581

Chapter 8 Functions of a Complex Variable

RULE V

If f(z) has a zero of first order at oo,then

: ~ MtTi :

,

\

r -6

Res [f (z), oo] = - lim zf (2). Z W '

If f (z) has a zero of second or higher order at oo, the residve at at is zero.

•.

RULE VI Res [f (z), ool = -Res

I

-f (llz), 0

[z:

.

The proof of Rule V is left as an exercise (Problem 8 following Section 8.18). To prove Rule VI, we write a-1 a-2 f ( z ) = . . . + a, z n + . . . + a ~ z + a o + (lzl > R). Z

Then for 0 <

lzl

+-+... z2

< R-',

Hence

and the rule follows. This result reduces the problem to evaluation of a residue at zero, to which Rules I through IV are applicable. EXAMPLE 1 We consider the integral

of Example 2 in the preceding section. There is no singularity outside the path other than oo,and at oo the function has a zero of order 3; hence the integral is zero. 71

EXAMPLE 2 1

5 "-

(Z lzl=2

+ 1 ) 4 ( ~-2 9)(z - 4) dz.

Here there is a fourth-order pole inside the path, at which evaluation of the residue is tedious. Outside the path there are first-order poles at f3 and 4 and a zero of order 7 at oo.Hence by Rule I the integral equals

4IF 582

Advanced Calculus. Fifth Edition

Let f (z) be analytic in a domain D. Then f '(z)/ f (z) is analytic in D except at the zeros of f (z). If an analytic branch of log f (z) is chosen in part of D [necessarily excluding the zeros of f (z)], then

f '(z)

d dz

- log f (z) = -

f (z)

'

For this reason the expression f '1f is termed the logarithmic derivative of f (z). Its value is demonstrated by the following theorem: THEOREM 32 Let f (z) be analytic in domain D. Let C be a simple closed path in D within which f (z) is analytic except for a finite number of poles and let f (z) # 0 on C. Then

where No is the total number of zeros of f inside C and N, is the total number of poles o f f inside C , zeros and poles being counted according to multiplicities. Proof. The logarithmic derivative f 'If has isolated singularities precisely at the zeros and poles of f. At a zero zo,

Hence the logarithmic derivative has a pole of first order, with residue N equal to the multiplicity of the zero. A similar analysis applies to each pole o f f , with N replaced by -N. The theorem then follows from the Cauchy residue theorem (Theorem 28), provided that we show that there are only a finite number of singularities. The poles of f are finite in number by assumption, and it is easily shown that there can be only a finite number of zeros. (See Section 6.6 of the book by Kaplan listed at the end of this chapter.) Thus the theorem is proved. Remarks.

Since

f '(z) dz = d log f (z) = d log w = d(1og Iwl + i arg w ) , f (z)

we have

.. 1

-,

*

dargw.

2ni C

C

C

Since w # 0 on C, log Iwl is continuous on C, and the first integral on the right is zero. The second measures the total charge in arg w , divided by 2n,as w traces the

Chapter 8 Functions of a Complex Variable

Figure 8.20

Argument principle.

path C,, the image of C, in the w-plane. Hence it also measures the number of times C, winds about the origin of the w-plane; in Fig. 8.20 the number is +2. (For more on this winding number see Problem 5 following Section 5.7). The statement 1 2n

- [increases in arg f ( z ) on path] = NO- N p

(8.82)

is known as the argument principle. This is of great value in finding zeros and poles of analytic functions. THEOREM 33 Let f (z) be analytic in domain D. Let C, be a piecewise smooth simple closed path in D, within which f (z) is analytic. If the function f (z) maps the curve C, in a one-to-one fashion onto a simple closed path C, in the w-plane and fl(z) # 0 inside C,, then f (z) maps the interior of C, in a one-to-one fashion onto the entire interior of C,.

Proof. We first remark that the Jacobian of the transformation from the xy-plane to the uv-plane is

Hence J > 0 inside C,. It then follows (see Section 5.14) that, as z traces C, in the positive direction, w = f (z) traces C, in the positive direction. Hence, if wo is a point interior to C, (Fig. 8.21), arg (w - wo) = arg [f (z) - wo] increases by precisely 2n.We apply the argument principle to the function f (z) - wo and, since this function has no poles inside C,, conclude that 1 No = - . 2 n = 1;

2n

that is f (z) - wo has precisely one zero zo inside C,. Every zo inside C, occurs as such a zero, for let f (20) = WO. Then wo cannot be outside C,, for then arg (w - wo)increases by 0 as C, is traced, so that f (z) - wo would have no zeros inside C,; also wo cannot be on C,, for then by the inverse mapping theorem (see Problem 9 following Section 8.7), f would map a neighborhood

583

584

Advanced Calculus, Fifth Edition

Figure 8.21

One-to-one mapping.

of zo (contained in the interior of C , ) onto a neighborhood of wo and hence f (21) = w, would be outside C , for some z, inside C , . Thus f establishes a one-to-one correspondence between the interiors of C, and C,.

COROLLARY Let f (z) be analytic in domain D, and map D, in a one-to-one fashion into the w plane. Then f '(z) # 0 in D,. Proof. If f '(zo) = 0, then choose a circle Cz about zo lying with its interior in D,. The function f (z) maps C , onto C , as in the above theorem and, as in the above proof, No = 1 for the function f (z) - wo. However, since f '(zo) = 0, f (z) - wo has a multiple root at 20, so that No > 2. This is a contradiction, hence f '(2) # 0 in D,. Remarks. As in Problem 9 following Section 8.7, the inverse z = g(w) of w = f (z) in Theorem 33 is analytic inside C , and gl(w) # 0. Thus f and its inverse provide one-to-one conformal mappings between the interiors of C , and C,. The condition f '(z) # 0 in D, is not sufficient to guarantee a one-to-one transformation throughout all of D,. For example, w = z2 satisfies this condition for z # 0, but the transformation is not one-to-one, since the inverse is the two-valued function z = f i .Sufficient conditions for a one-to-one mapping, supplementing Theorem 33, are given in Section 8.20.

The theory of analytic functions provides a simple proof for the familiar rules for ' partial fraction expansions. Let

be given. We assume that n c m , so that f is a proper fraction. We also assume that the numerator and denominator have no common zeros. Let zl,z2, . . . , Z N be the distinct zeros of the denominator (no repetitions); these are the poles of f (2).

Chapter 8 Functions of a Complex Variable At z l , f ( z ) has a Laurent expansion:

Here p l ( z ) is the principal part o f f ( z ) at z l and g l ( z ) is analytic at 21; k l is the order of the pole at z l . Similar expressions hold at the other poles. The partial fraction expansion of f ( z ) is now simply the identity To justify this, we let Now f ( z ) - pl ( z ) has a removable singularity at z l , while P ~ ( Z .) .,. , P N ( 2 ) are analytic at z l . Hence F ( z ) has a removable singularity at z l . In general, F ( z ) has only removable singularities for finite z. At oo,f ( z ) , p l ( z ) , . . . , p N ( z )all have zeros; hence F ( z ) has a zero at oo.But F ( z ) is a rational function with no poles. Hence F ( z ) must be a polynomial. Thus F ( z ) has a pole at oo unless F is constant; since we know that F = 0 at oo, F must be a constant, namely, zero. This proves (8.85). If f ( z ) has only simple poles, the principal part at each pole z is simply A / ( z - z,), where A j is the residue of f at z j , and hence in this case

is the partial fraction expansion. If we write

then Rule I11 can be applied:

At a multiple pole z j of order k we can write

f (z)=

1 ( 2 - z,Ik

The principal part p,(z) is then

Hence p j ( z ) can be found without knowledge of the other poles.

EXAMPLE 1

Y

€

585

586

Advanced Calculus, Fifth Edition There are simple poles at 0,- 1 , -3. By (8.87),

with A = z2

+ 1 , B = z 3 + 4 z 2 + 3 z , B' = 3z2 + 82 + 3 we find

EXAMPLE 2

f (2)=

z ( Z - 112(z3+ z + 1)-

At the pole z = 1 we write

Since @ ( I )= !, Q'(1) = -

i,the principal part at 1 is

+ +

The cubic z3 z 1 has one real root zl and two complex roots z2,z3. These are all simple. Hence we can write

where A ( z ) = z , B ( z ) = ( z to z2 z 7 , by the relation z3

+ +

+ z + 1). At the poles z l , 22, z3, B 1 ( z )reduces + z + 1 = 0. e>

PROBLEMS 1. Evaluate the following integrals on the paths given: a) f ( z -

i$-N

(17-1 = 2)

b)

f

' ~ d z (kt = 3)

c)

f ydz (lzl = 1)

1

$ 3

e,

dZ (lzl = 2) ' ( z + 1)4(z+3)

f,

f ( z + 1$z+3)

(121 = 4)

2z+2 dz (lzl = 2) g, + z 2 + 2 z + 2

h,

j z 3 3- z3 2z -26+z z+-13

dz (121 = 2)

(111 = 4)

+

2. Expand each of the following in partial fractions: 1

a) ;r_4 z2

c, 7-z

GiibF 1 (z-1)'(z4+z+l)

z+ 1

b, i z - l ) ( z - 2 ) ( z - 3 d) f)

h,

* 1

'rn=c0+c,z+-..+cnzn,

1 (z2+1)(z2+2z+2)

Chapter 8 Functions of a Complex Variable 3. Prove the fundamental theorem of algebra: Every polynomial of degree at least 1 has a zero. [Hint: Show that Res [ f l ( z ) / f ( z ) . m ] is not 0, but is in fact minus the degree n of the polynomial f (z). Then use Theorem 32 to show that f has n zeros.] 4. Prove Theorem 30. 5. Formulate and prove Theorem 32 for integration around the boundary B of a region R in D, bounded by simple closed curves C I ,. . . , Ck. 6. Prove that under the hypotheses of Theorem 32, if g ( z ) is analytic in D and within C, then

where 2 ; . . . . , z; are the zeros o f f , and 7;. . . . , Z; are the poles of f inside C, repeated according to multiplicity. 7. Extend Rule IV of Section 8.15 to the case in which A ( z ) has a first-order zero at zo and B ( z ) has a second-order zero. 8. Prove Rule V of Section 8.16.

A variety of real definite integrals between special limits can be evaluated with the . aid of residues. For example, an integral . ,
b2*

where R is a rational function of sin8 and cos8, is converted to a complex line integral by the substitution:

- e-i@

sin z = the path of integration is the circle

EXAMPLE 1 :J

since -2

2i

1z.l = 1.

&dB. The substitution reduces this to

+ a is the only root of the denominator inside the circle. Accordingly,

587

588

4-d

Advanced Calculus, Fifth Edition The substitution can be summarized in the one rule:

. *' '

.*

>*-

'- .

.

The complex integral can be evaluated by residues, provided that R has no poleBon the circle 1 z 1 = 1 . A second example is provided by integrals of the type

We illustrate the procedure with an example and formulate a general principle below.

l-",5.

EXAMPLE 2 This integral can be regarded as a line integral of f (z) = l/(z4 1) along the real axis. The path is not closed (unless one adjoins oo),but we show that it acts like a closed path "enclosing" the upper half-plane, so that the integral along the path equals the sum of the residues in the upper half-plane. To establish this, we consider the integral of f ( 2 ) on the semicircular path C R shown in Fig. 8.22. When R is sufficiently large, the path encloses the two poles: zl = exp(ain), z2 = exp(:in) o f f (z). Hence

+

As R increases, the integral on C Rcannot change, since it always equals the sum of the residues times 2n. Hence

0 5 t9 5 n. The limit of the first term is the where D R is the semicircle z = Reie, integral desired (since the limits at +oo and -00 exist separately, as required; cf.

Figure 8.22

Evaluation of [_bO, f ( x ) dx by residues.

Chapter 8 Functions of a Complex Variable Section 4.1). The limit of the second term is 0, since (z4 1 I 1, SO that on DR,

+ +

1214

= 1z41 = 1z4

+ 1 - 11 5

and

Accordingly,

By Rule I11 the sum of the residues is

,. .i.

and hence

We now formulate the general principle: THEOREM 34 Let f ( z ) be analytic in a domain D that includes the real axis and all of the half-plane y > 0 except for a finite number of points. If

lim

R-+ w

lr

f ( ~ e ' ' ) ~ e do '' =0

and

exists, then 00

f ( x )dx = 27ri {sum of residues of f ( 2 ) in the upper half-plane].

(8.92)

The proof is a repetition of the reasoning used in the above example. It is of interest to note that even when the integral (8.91) fails to exist as improper integral, condition (8.90) implies that the principal value

exists. (See Section 6.24.) *-

589

-4

&

A-dvanced Calculus, Fifth Edition

In order to apply Theorem 33 it is necessary to have simple criteria to gdarafitee that (8.90) holds. We list two such criteria here:

I. If f (z) is rational and has a zero of order greater than 1 at oo,then (8.90) holds. For when

)zl is sufficiently large,

so that zf (z) has a zero at infinity. Now, by (8.1 l),

I ln

In

Izf (z)I do,

f ( ~ e " ) R e ' @ d9 ~I

so that the integral must converge to 0 as R -+ m. 11. If g(z) is rational and has a zero of order 1 or greater at oo,then (8.90) holds for f (z) = em' 0.

1

For a proof and further criteria we refer to page 1 15 of the treatise of Whittaker and Watson listed at the end of the chapter. Rule I1 makes possible the evaluation of the Fourier integral (Section 7.18),

LO)

[_ g(x) cos mx w

g(x)em'xdx =

dx

+i

w

,-

provided that both real integrals exist.

EXAMPLE 3 The integrals x cos x

dx

and

Srn -,+

x sin x -d x X2 1

both exist by the corollary to Theorem 5 1 of Section 6.22. Hence

Taking real and imaginary parts, we find x cos x

x sinx

.

x

dx = -. e

Since the first integral is an integral of an odd function, the value of 0 could be predicted. •

PROBLEMS 1. Evaluate the following integrals:

Chapter 8 Functions of a Complex Variable 2. Evaluate the following integrals: a)

1-z

C)

S-,

x2

+: , +

1

aJ

b,

dx

3. Evaluate the following integrals: a) -j: dx C)

10ao x'

sin x

1

:-S ( x 2 + 1 ) ( x 2 + 4 ) d x d) Si (x2 ,)2 d x

dx

: x::z

b) aJ

dx

d,

dx

x cos 3 x d x (x~+I)~

Sr

d x = in. [Hint: Let C be a path formed of the semicircular paths = r and DR: I z I = R, where 0 5 8 e n and O t r c R , plus the intervals Dr: lzl -R 5 x 5 -r, r 5 x 5 R on the real axis. Show that

4. Prove that

Use this result and Rule I1 to conclude that r ,iz

- dz = 0.

lii",S, T d z = - x i .

Accordingly, lim

R+w

[ f$ lim

r+O

d l ] = lim R+w

( [/_: lim

r+O

dx

+

lR

d x ] ] - ni.

C

The left-hand side is zero by the Cauchy integral theorem. Show that the imaginary part $ ; d x - n.] of the right-hand side has the value 2

%

As pointed out in Section 8.6, a function w = f ( z ) can be interpreted as a transformation or mapping from the z plane to the w plane. The term "mapping" is really justified only when the correspondence between z values and w values is one-to-one, i.e., to each z of a domain D, corresponds just one w = f ( z ) of a domain D, and conversely. The domain D, is then a distorted picture or "image" of the domain D,; circles in D, correspond to closed curves in D,, triangles in D, to curvilinear triangles in D,, as illustrated in Fig. 8.23.

Figure 8.23

Conformal mapping.

591

592

Advanced Calculus, Fifth Edition If f (z) is analytic, such a mapping has an additional property, that of being conformal. A mapping from D, to D, is termed conformal if to each pair of curves in Dz intersecting at angle a there corresponds a pair of curves in D, intersecting at angle a . The mapping is termed conformal and sense-preserving if the angles are equal and have the same sign, as illustrated in Fig. 8.23. THEOREM 35 Let w = f (z) be analytic in the domain D;, and map D,in a oneto-one fashion on a domain D,. If f'(z) # 0 in D,, then f (z) is conformal and sense-preserving.

+

Proof. Let z(t) = x(t) iy(t) be parametric equations of a smooth curve through zo in DZ.By proper choice of parameter (e.g., by using arc length) we can ensure that the tangent vector dz dx dy - -- - + i dt dt dt is not 0 at zb. The given curve corresponds to a curve w = w(t) in the w plane, with tangent vector dw - d-wdz dt dz d t as in Section 8.2. Hence dw dw dz arg - = arg - arg -. dt dz dr This equation asserts that, at wo = f (zo), the argument of the tangent vector differs from that of d z l d t by the angle arg f '(zo), which is independent of the particular curve chosen through zo. Accordingly, as the direction of the curve through zo is varied, the direction of the corresponding curve through wo must vary through the same angle (in magnitude and sign). The theorem is thus established.

+

Conversely, it can be shown that all conformal and sense-preserving maps w = f (2) are given by analytic functions; more explicitly, if u and v have continuous first partial derivatives in D, and J = a(u, v)/a(x, y) # 0 in D,, then the fact that the mapping by w = u i v = f (z) is conformal and sense-preserving implies that ux = v,, u y = -v,, so that w is analytic (see Section 2.3 of the book by Ahlfors listed at the end of the chapter). From this geometric characterization it is also clear that the inverse of a one-to-one conformal, sense-preserving mapping has the same property and is itselfanalytic. (See Problem 9 following Section 8.7.) I f f '(zo) = 0 at a point zo of D,, then arg f '(zo) has no meaning and the argument above breaks down. One can in fact show that conformality breaks down at zo;indeed, the transformation is not one-to-one in any neighborhood of zo by the Corollary to Theorem 33 in Section 8.17; see also Problem 14 following Section 8.21. In practice, the term "conformal" is used loosely to mean "conformal and sensepreserving"; that will be done here. It should be noted that a reflection, such as the mapping w = Z, is conformal but sense-reversing.

+

Testsfor one-to-one-ness. For the applications of conformal mapping, it is crucial that the mapping be one-to-one in the domain chosen. In most examples, the mapping will also be defined and continuous on the boundary of the domain; failure of oneto-one-ness on the boundary is less serious.

Chapter 8 Functions of a Complex Variable By the Corollary to Theorem 33 in Section 8.17, if f ' ( z ) = 0 at some point of the domain. then the mapping cannot be one-to-one. One should therefore verify that f ' ( 2 )# 0 in the domain as a first step. Even if this is satisfied, the mapping need not be one-to-one, and one must apply additional tests. The following are useful tests for one-to-one-ness:

I. Explicit fortnula for the inverse function.

If an explicit formula for the inverse function z = z ( w )is available and one can show that, by this formula, there is at most one z in D, for each w , then the mapping must be one-to-one. For example, w = z' is one-to-one in the first quadrant of the z plane, for to each w there is at most one square root f i in the quadrant. As 7 varies over D,,w varies over D,,: the upper half-plane v > 0, as will be seen. 11. Analysis of level curves of u and v. One can verify one-to-one-ness and at the same time obtain a very clear picture of the mapping by plotting the level curves: u ( x , y ) = c l , v ( x , y ) = c2. If for arbitrary choice of c l , c2 the loci u = c l , v = c2 intersect at most once in D,,then the mapping from DZto the uv plane is one-to-one. 111. One-to-one-ness on the boundary. This is the test formulated in Theorem 33 in Section 8.17: if f ( z ) is analytic in C plus interior D, and one-to-one on C, then f ( z ) is one-to-one on D,. As one can show, it is sufficient that f be continuous on C plus interior, and analytic on the interior; in various important examples this is the case, analyticity being violated at one or more points on the boundary. One can also reason thus: if one can show that, 8 for every simple closed curve C' approximating the boundary sufficiently closely, f is one-to-one on C', then f must be one-to-one in D:; for if f fails to be one-to-one in D,,then f ( 2 1 ) = f ( z 2 ) for two points zl , zz of D,. A curve C' sufficiently close to the boundary would enclose both 7 and z2, SO that the failure of one-to-one-ness would have to reveal itself on C'. Other natural extensions of the principle will be pointed out in the examples to follow. IV. Real or imagitzary part maxin~aand rninima on the boundary. Let f ( z ) be analytic in domain D,,bounded by the simple closed curve C . Let u = Re [f ( z ) ]be continuous in D, plus C. Ifu hasprecisely one relative maximum and one relative minimuln on C, then w = f ( 2 ) is one-to-one in D,. A similar criterion holds for v = Im [f ( z ) ] .Furthermore, the same conclusion holds for mappings: u = u ( x , y ) , v = v ( x , y ) by nonanalytic functions, provided the Jacobian J = a(u, v)/a(.u, y ) is always positive (or always negative) in D;. For a proof see the paper by Kaplan listed at the end of the chapter. V. Let f ( z ) = u i v be analytic in D; and let DZbe convex; that is, for each pair of points zl and z2 of D,,let the line segment from z l to z2 lie in D,. If real constants a , b can be found such that

+

then f is one-to-one in D,(see Problem 15 below, following Section 8.21).

4 594

Advanced Calculus, Fifth Edition

VI. Let f ( 2 ) = u + i v be analytic in D, and let D,be the half-plane: y > 0. If a complex constant c and a real number a can be found, such that Re [c(z - a )f '(z)] > 0

1

in D,, then f is one-to-one in DZ(see Problem 16 following Section 8.21).

EXAMPLE 1

(Translations) The general form is w =z

+ a + bi

(a, b real constants).

Each point z is displaced through the vector a

EXAMPLE 2

(Rotation-Stretchings)

+ bi, as in Fig. 8.24.

The general form is

w = A ~ ; @ z ( A , a real constants, A > 0).

If one writes z = re'e, w = pei@,so that p , c$ are polar coordinates in the w plane, then one has

Thus distances from the origin are stretched in the ratio A to 1, while all figures are • rotated about the origin through the angle a (Fig. 8.25).

Figure 8.24

Translation.

Figure 8.25

Rotation-stretching.

Chapter 8 Functions of a Complex Variable z and w

Figure 8.26

EXAMPLE 3

The transformation w = f .

(The General Linear Integral Transformation) w = az

+b

( a , b complex constants).

(8.94)

This is equivalent to a rotation-stretching, as in Example 2, with a = Aeia, followed by a translation through the vector b. EXAMPLE 4

(The Reciprocal Transformation)

In polar coordinates one has

Thus this transformation involves both a reflection in the real axis and "inversion" in the circle of radius 1 about the origin (Fig. 8.26). Figures outside the circle correspond to smaller ones inside. It can be shown that circles (including straight lines, as "circles through 00")correspond to circles (Problem 5). Thus the line x = 1 becomes the circle

EXAMPLE 5

(The General Linear Fractional Transformation) w=-

az+b cz + d

( a , b, c, d complex constants),

595

S96

Advanced Calculus, Fifth Edition If ad - bc were equal to 0, w would reduce to a constant; hence this is ruled out. Examples 1, 2, 3, and 4 are special cases of (8.95). Moreover the general transformation (8.95) is equivalent to a succession of transformations,

of the types of Examples 3 and 4; if c = 0, (8.95) is already of type (8.94). The transformation (8.95) is analytic except for z = - d / c . It is one-to-one, for (8.95) can be solved for z to give the inverse:

which is single-valued; w has a pole at z = -d/c and z has a pole at w = a l c ; in other words, z = - d / c corresponds to w = oo and z = oo corresponds to w = a l c . If one includes these values, then the transformation (8.95) is a one-to-one transformation of the extendedplane on itselj When c = 0, z = oo corresponds to w = oo. Since the transformations (8.96) all have the property of mapping circles (including straight lines) on circles, the general transformation (8.95) has also this property. By considering special domains bounded by circles and lines, one obtains a variety of interesting one-to-one mappings. The following three are important cases: *

EXAMPLE 6

(Unit Circle on Unit Circle) The transformations

'

w = eia- Z" (areal, Izol < 1) 1 - 202

all map lz 1 5 1 on Iw 1 5 1, and every linear fractional (or even one-to-one conformal) transformation of J z J5 1 on lwl _< 1 has this form. (See Problem 11). EXAMPLE 7

(Half-Plane on Half-Plane) The transformations

w =a' - ( a ,b, c, d real and ad - bc > 0) cz d all map Im ( z ) 2 0 on Im (w) 2 0, and every linear fractional transformation of Im (z) 2 0 on Im (w) 2 0 has this form (see Problem 12).

+

+

EXAMPLE 8

(Half-Plane on Unit Circle) The transformations w=

,ia

Z - Zo -

z - 20

(areal, Im (zo) > 0)

(8.100)

all map Im (z) 2 0 on Iw 1 5 1, and every linear fractional transformation of Im (z) 2 0 on 1 w 1 ( 1 has this form (see Problem 13). EXAMPLE 9

(Squaring) The equation is w = zL.

Here the transformation is not one-to-one in the whole z plane, for the inverse is z = f i and has two values for each w. In polar coordinates one has

Chapter 8 Functions of a Complex Variable

Figure 837

The transformation w = z2.

Thus each sector in the z plane with vertex at the origin and angle a! is mapped onto a sector in the w plane with vertex at w = 0 and angle 2a. In particular the half-plane D,: Im (z) > 0 is mapped on the w plane minus the positive real axis, as shown in Fig. 8.27. Since the two square roots of w are negatives of each other, each w has at most one square root in D,, so that the mapping is one-to-one. The structure of the level curves

u = x 2 - y 2 = const,

v

= 2 x y = const,

shown in Fig. 8.27, also reveals the one-to-one-ness, in accordance with test I1 of Section 8.20. Test V gives the inequality: a x -by > 0.Such an inequality is satisfied in each half-plane bounded by a line through the origin; hence w = z2 is one-to-one in every such half-plane. It should be noted that d w l d z = 0 for z = 0 and that the mapping is not conformal at this point; the angles between curves are doubled. EXAMPLE 10

(Other Integral Powers) The equation is

Here the inverse z = w ' l n is n-valued. In polar coordinates

so that each sector is expanded by a factor of n. One obtains a one-to-one mapping by restricting to a sector: 0 < arg z < 2rr/n, as illustrated in Fig. 8.28. The one-toone-ness follows from the explicit formula:

for the inverse function, or by one of the other tests. Actually n need not be an integer; for n any positive real number, similar results hold.

EXAMPLE 11 (The Exponential Transformation) The equation is w = ez.

597

598

Advanced Calculus, Fifth Edition

I

r. = 0

I

Figure 8.28

The transformation w = zn.

Figure 8.29

The transformation w = e .

Here the inverse z = Log w has infinitely many values. In polar coordinates:

This shows that lines x = const in the z plane become circles p = const in the w plane, while lines y = const become rays: 4 = const. One obtains a one-to-one mapping by restricting z to a strip: -n c y c n, which is then mapped onto the w plane minus the negative real axis, as shown in Fig. 8.29.

599

Chapter 8 Functions of a Complex Variable

Figure 8.30

The transformation w = sin z.

EXAMPLE 12 (Sine Transformation) The equation is

w = sinz. Here the inverse 1

z = sin-' w = log (i w 1

+ Jw)

is infinitely multiple-valued. One obtains a one-to-one mapping by restricting z to the strip -in < x < in,as the structure of the level curves of Fig. 8.30 reveals. One can also verify that v = cos x sinh y has precisely one maximum and one minimum on each rectangle with vertices (fin, fk); hence by test IV we conclude that the mapping is one-to-one inside each such rectangle, and accordingly in the whole strip. Test V is also applicable, with a = 1 and b = 0.

EXAMPLE 13 (A Rational Transformation) The equation is 1 z The inverse is the two-valued solution of the quadratic equation

w=z+-.

z2 - z w + l = O .

i;

The mapping is one-to-one in the domain: 1z.1 > 1 , as Fig. 8.31 reveals (see also Problem 3 following Section 8.26).

-

Advanced Calculus, Fifth Edition

Figure 8.31

The transformation w = Z + (112).

A variety of other mappings can be obtained by applying first one of the above mappings and then a second one, since the composite of two conformal mappings is conformal. Also the inverse of a conformal transformation is analytic and one-to-one in the appropriate domain (see Section 8.20). Section 8.27 discusses general classes of mappings.

PROBLEMS 1. Determine the images of the circle lz - 1I = 1 and the line y = 1 under the following transformations: b) w = z + 3 i - 1 c)w=2iz a) w = 22 1 d)w=, e ) w = *z - I f ) w = z 2. 2. For each of the following transformations verify that the transformation is one-to-one in the given domain, determine the corresponding domain in the w-plane, and plot the level curves of u and v: a) w =

JZ= ,/7eieI2,

- n < 0 < n.

w = y1, 1 < x < 2 . d) w = Logz, Im(z) > 0. C)

9,

e) w LO^ Im(z) z 0. [Hint: Use test VI, with c = i and a = 0.1 f) w =z Im(z) > 0. [Hint: Use test VI, with c = -i and a = 0.1

+ i,

g) w = z

,,

-1

lzl > 1 h) w =Log & - 2Logz, Im(z) > 0 [Hint: Use test V, with a = 0 and b = - 1.1

601

Chapter 8 Functions of a Complex Variable

3. Verify the mapping onto the domain shown and the level curves of u and v for the transformations of a) Fig. 8.27 b) Fig. 8.28 c) Fig. 8.29 d) Fig. 8.30 e) Fig. 8.3 1. 4. By combining particular transformations given above determine a one-to-one conformal mapping of a) the quadrant x > 0, y > 0 onto the domain Iw 1 < 1

b) the sector 0 < 8 < i n on the quadrant u > 0, v > 0 c) the half-plane y > 0 on the strip 0 < v c n d) the half-strip - i n < x c i n , y > 0 on the quadrant u z 0, v z 0 e) thedomainr > 1 , 0 < 9 < n o n t h e s t r i p O c v < n f ) the strip 1 c x y < 2 on the half-plane v > 0 g) the half-plane x y 1 > 0 on the quadrant u > 0, v > 0. 5. a) Show that the equation of an arbitrary circle or straight line can be written in the form

+ + +

+ + 61 + c = O

a z ~ bz

( a , c real).

b) Using the result of a), show that circles or lines become circles or lines under the transformation w = l/z. 6. If w = f (z) is a linear fractional function (8.95), show that f '(2) # 0. 7. A point where f '(z) = 0 is called a critical point of the analytic function f (z). Locate the critical points of the following functions and show that none lies within the domain of the corresponding mapping in Examples 11, 12, 13: a) w = ez b) w = sin z C) W = Z + ~1 .

8. a) Show that the determinant of coefficients of the inverse (8.97) of the transformation (8.95) is not zero. Thus the inverse of a linear fractional transformation is linear fractional. b) Show that, if

are linear fractional transformations: wl = f ( z ) , w2 = g(wl), then wz = g[ f (z)] is linear fractional. Thus a linear fractional transformationfollowed by a linear fractional transformation gives rise to a linear fractional transformation. 9. The cross-ratio of four complex numbers zl, z2,23,24, denoted by [zl, 22, z3, Z4], is defined as follows:

This is meaningful, provided at least three of the four numbers are distinct; if z l = 24 or ~2 = 23 it has the value oo; the expression remains meaningful if one, but only one, of the numbers is oo. a) Rove that, if Z , , Z ~ ,are Z ~distinct, then w = [z, 21,z2,231 defines a linear fractional function of z.

I

602

Advanced Calculus, Fifth Edition b) Prove that, if w = f ( z ) is a linear fractional function and wl = f ( z l ) ,w2 = f (Q), " w3 = f ( ~ 3 1w4 , = f ( ~ 4 ) then .

C)

provided at least three of z l , z2,z3, z4 are distinct. Thus the cross-ratio is invariant under a linear fractional transformation. Let zl , z2, z3 be distinct and let W I , w2, wg be distinct. Prove that there is one and only one linear fractional transformation w = f ( z ) such that f ( z l ) = W I ,f ( z 2 ) = w2. f ( z 3 )= w3, namely the transformation defined by the equation

[z. Z I ,z2. 231 = [ w , W I . w2, w31. d) Using the result of c) find linear fractional transformations taking each of the following triples of z values into the corresponding w's: i ) z = 1 0 w = 0, i, 1; (ii) z = 0 , 00, 1; w = w , O , i. 10. Let C be a circle of radius a and center Q . Points P , P' are said to be inverses of each other with respect to C if P is on the segment QP' or P' is on Q P and Q P . QP' = a2. Q and oo are also considered as a pair of inverse points. If C is a straight line, points P , P' symmetric with respect to C are called inverses of each other. a ) Prove: P , P' are inverses with respect to C,if and only if every circle through P, P' cuts C at right angles. b) Prove: If P , P' are inverses with respect to C and a linear fractional transformation is applied, taking P to P I , P' to P;, C to C I ,then P I , Pi are inverses with respect to C I. [Hint: Use a), noting that circles become circles and right angles become right angles under a linear fractional transformation.] 11. a ) Let w = f ( 2 ) be a linear fractional transformation mapping 171. 5 1 on I w l 5 1. Prove that f ( z )has the form (8.98). [Hint: Let zo map on w = 0. By Problem 10 b) 1/20 must map on w = oo.The point z = 1 must map on a point w = eiB. Take zl = 1, z 2 = zo, z3 = 1/To in 9 c).] b) Prove that every transformation (8.98) maps lz 1 <: 1 on I w 1 5 1. 12. Prove that every linear fractional transformation of Im ( z ) _> 0 onto Im ( w ) 2 0 has the form (8.99) and that every transformation (8.99) maps Im ( z ) 2 0 onto Im ( w ) 2 0. 13. Prove that every linear fractional transformation of Im ( z )1.0 on I w 1 5 1has form (8.100) and that every transformation (8.100) maps Im ( z ) 1.0 onto I w 1 5 1 . 14. Let w = f ( 2 ) be analytic at zo and let f '(zo) = 0. . . . , f ( " ' ( ~ 0 )= 0 , f ("+l)(zo)# 0 . Prove that angles between curves intersecting at zo are multiplied by n + 1 under the transformation w = f ( z ) .[Hint: Near zo, w = wo cn+1(t - zoy" C ~ + Z(Z . ., where cn+l # 0. A curve starting at zo at t = 0 with tangent vector (complex number) a # 0 can be represented as z = zo a t c t , 0 5 t 5 t1, where 6 = ~ ( t -+ ) 0 as t --+ O+. Show that the corresponding curve in the VJ plane can be represented as w = wo + ~,+~a"+lt"+' cl(t)tn+'. Show that ( w - w o ) / t n f lhas as limit, for t + 0+, a tangent vector to the curve at wo, and that this tangent vector has argument arg cn+l +(n + 1) arg a . From this conclude that two curves in the z plane forming angle cr at zo become two curves in the w plane forming angle ( n l)a at wo.] 15. Prove the validity of test V. [Hint: Let c = a + ib. Let zl and zz = 2 , reia be distinct points of D. Then

+

+

+.

+ +

+

+

C [f

(9) - f ( z I ) ]= c

1:'

f ' ( z )d z = eia

+

ir

~f'(L)d t ,

Chapter 8 Functions of a Complex Variable where z = zl + teIu on the straight-line path of integration. Show that the real part of the last integral is positive, so that f ( z 2 ) # f (zl) Cf. F. Herzog and G. Piranian, Proceedings of the American Mathematical Society, Vol. 2 (1951), pp. 625-633.1 16. Prove the validity of test VI. [Hint: Set zl = Log ( z - a ) and show that the function wl = f [ z ( z l ) ] = g(z1) is analytic and satisfies the inequality: Re [cgl(zl)]> 0 in the domain: 0 < Im ( z l ) c a . Now apply test V.]

The following problem, known as the Dirichlet problem, arises in a variety of situations in fluid dynamics, electric field theory, heat conduction, and elasticity: given a domain D, to$nd afunction u ( x , y ) harmonic in D and having given values on the boundary of D. The statement is somewhat loose as to the boundary values. It will be seen immediately how it can be made more precise. Let D be a simply connected domain bounded by a simple closed curve C, as in Fig. 8.32. Then assignment of boundary values is achieved by giving a function h ( z ) for z on C and requiring that u = h on C. If h is continuous, the natural formulation of the problem is to require that u ( x , y) be harmonic in D, continuous in D plus C, and equal to h on C. If h is piecewise continuous, it is natural to require that u be harmonic in D, continuous in D plus C , except where h is discontinuous, and equal to h except at the points of discontinuity. If C is a circle, lzl = R, the problem is solved by the Poisson integral formula of Section 5.17. For example, if R = 1, h can be written as h(8),a function of period. 21r, and

defines a function harmonic in lzI < 1 . If h ( 8 ) is continuous and we define u ( l , 8 )to be h(8),then u can be shown to be continuous for ( z (5 1 and therefore satisfies all cbnditions. If h(0) is discontinuous, the same procedure is successful and (8.101) again provides a solution. While the formula (8.101) does provide a solution of the problem, we must also ask: Is this the only solution? The answer is yes when h is continuous for lzl = 1 ; the answer is no if h has discontinuities. In the latter case (8.101)provides a solution

Figure 8.32

Dirichlet problem, .

603

604

Advanced Calculus, Fifth Edition u(x, y ) which is bounded for (zI < 1, and it can be shown that this is the only bounded solution. With the supplementary requirement that the solution be bounded, (8.101) provides the one and only solution. Now let D be a domain bounded by C as in Fig. 8.32 and let us suppose that a one-to-one conformal mapping: zl = f (z) of D onto the circular domain lz I < 1 has been found and that this mapping is also continuous and one-to-one on D plus C, taking C to (2, I = 1. By assigning to each point z l on lzl I = 1 the value of h at the corresponding point on C, we obtain boundary values h 1 (7-1)on J z1 ( = 1. Let u(z1) solve the Dirichlet problem for lzl 1 < 1 with these boundary values. Then u[f (z)] is harmonic in D and solves the given Dirichlet problem in D. For ~ ( z Ican ) be written as Re [F(z,)] where F is analytic and u[f ( z ) ] = Re ( F[f (z)]); i.e., u [f ( z ) ] is the real part of an analytic function in D . Hence u is harmonic in D. Since continuous functions of continuous functions are continuous, u [f (z)] will have the proper behavior on the boundary C and hence does solve the problem. Accordingly, conformal mapping appears as a powerful tool for solution of the Dirichlet problem. For any domain which can be mapped conformally and one-toone on the circle lzl < 1, the problem is explicitly solved by (8.101). It can be shown that every simply connected domain D can be mapped in a one-to-one conformal manner on the circular domain: 121 < 1 , provided D does not consist of the entire z-plane. Furthermore, if D is bounded by a simple closed curve C, the mapping can always be dejned on C so as to remain continuous and one-to-one. Proofs of these theorems appear in the following texts listed at the end of the chapter: Ahlfors (pp. 229-232), Bieberbach (1967, pp. 67-71), and Heins (pp. 130-134). For proofs of the stated properties of the Poisson integral formula, see Ahlfors (pp. 166-168) and Nevanlinna (pp. 22-24). There now remains the question of how to map a particular simply connected domain D onto a circular domain. Further information on this is given in Section 8.27 below. We shall not consider extension of the theory to multiply connected domains.

PROBLEM FOR THE HALF-PLANE 8.23 DIRICHLET

I

Since the transformation ZI

-i

iz = ZI + i

(8.102)

maps the domain lzl 1 < 1 on the half-plane Im (z) > 0, the Dirichlet problem for the half-plane is reducible to that for the circle as above. However, it is simpler to treat the half-plane by itself. We shall develop the equivalent of (8.101) for the half-plane. Accordingly, if a domain D can be mapped on the half-plane, the Dirichlet problem for D will be immediately solved. We consider first several examples.

EXAMPLE 1 u(x, y ) harmonic for y > 0;u = rr for y = 0, x < 0;u = 0 for y = 0, x > 0 as shown in Fig. 8.33. The function

Chapter 8 Functions of a Complex Variable

Y,

13= arg z

Figure 8.33

*.r

Example 1.

* Y

Figure 8.34

Example 3.

clearly satisfies all conditions. It is harmonic in the upper half-plane, since it is continuous on the boundary y = 0, except for x = 0, and has the correct boundary values. Furthermore 0 < u c IT in the upper half-plane, so that the solution is bounded. EXAMPLE 2 u(x, y) harmonic for y > 0; u = h = const, for y = 0, x > 0; u = 0 for y = 0, x < 0. The solution is obtained as in Example 1:

=

-

)

Osargz_ca.

This function is again harmonic and has the proper boundary values.

a

EXAMPLE 3 u(x, y) harmonic for y > 0; u = h = const, for y = 0, x > X I ; u = 0 for y = 0, x < xl; see Fig. 8.34. A translation reduces this to Example 2:

o

u =h ( ~ - ~ ~ ~ ( : - ~ ~ ~ 5 ) ,arg(z-zI) 5

n;

zl

= xl+0i.

EXAMPLE 4 u(x, y) harmonic for y > 0; u = h = const, for y = 0, xl < .x < x2; 14 = 0 for y = 0, x < x l and x > x2; see Fig. 8.35. The solution is obtained by subtracting two solutions of the type of Example 3: u=h

(

1-

arg(z-zl))-h(lIT

arg (z - 7.2) IT

h h Z-7.2 = -[arg(z - 22) - arg(2 - zo1 = - arg -. n IT Z-ZI

(8.103)

605

606

Advanced Calculus, Fifth Edition

Figure 8.35

Example 4.

Figure 8.36

Example 5.

The result has an interesting geometric interpretation, for

where

+ is the angle shown in Fig. 8.35. Accordingly,

One can directly verify the boundary values an the figure. Examples 1, 2, and 3 can be regarded as limiting cases of Example 4. In Example 1, xl = -m, x2 = 0, and h = n, so that $J becomes the angle 8 , as shown in Fig. 8.33, and u = $J = 8. In Example 3, x2 = +m; the angle $J is shown in Fig. 8.34.

Remark.

EXAMPLE 5 u(x, y) harmonic for y > 0; u = ho = const, for y = 0, x < xo; u=hlfory=O,x~~x~xl;u=h~fory=O < x, x< lx z ; ... ; u = h , f o r y = 0, x,-1 < x < x,; u = h,+l for y = 0, x > x, as in Fig. 8.36. The solution is obtained by addition of solutions of problems like those in the previous examples: n

ho arg ( z - 2 0 )

+h

2-21

1

- Zo + . . . + h,

arg Z

7

Z

- Zn

arg Z - Zn-1

(8.104)

Chapter 8 Functions of a Complex Variable

607

The angles $0, . . . , are shown in Fig. 8.36. The sum of these angles is n ; therefore u is the weighted mean of the numbers ho, hl, . . . , hn+l and hence h' p u 5 h", where h' is the smallest of these numbers and h" is the largest. Now let ho = 0, h,+l = 0. Let z be fixed in the upper half-plane and let t be a variable point on the x axis; write g(t>= arg [z - (t

+ Oi

+ Oi)],

where the angle is always taken between 0 and n. Then (8.104) can be written thus:

This formula suggests passage to the limit: n + oo; the expression suggests an integral. In fact, the expression can be interpreted (with the aid of the Mean Value theorem) as a sum 1

= -[h(tt)gl(t;)Alt

+ +

. h(tn)gl(t,*)Antl n which (under appropriate assumptions) converges as n + oo to an integral: u

Now g(t) = arg [z - (t

+ Oi)] = arctan -,x -Y t

so that

We are thus led to the formula:

as the expression for a harmonic function in the upper half-plane with boundary values h(t) for z = t Oi, a < t < B , and u = 0 otherwise on the boundary. We get complete generality by letting a = -00, B = +oo:

+

This integral is easily shown to converge, provided h is piecewise continuous and bounded. The formula (8.107) is precisely the Poisson integral formula for the halfplane. In fact a change of variables (Problem 6 below) transforms (8.101) into (8.107). Just as important as the general formula (8.107) is the formula (8.104), which can be considered as the rectangular sum for approximate evaluation of the integral (8.107).

a

& i

Advanced Calculus, Fifth Edition

Figure 8.37

Example 6.

EXAMPLE 6 u ( x , y ) harmonic in the half-strip of Fig. 8.37, with the boundary values shown. One seeks the mapping of this half-strip onto a half-plane and finds it to be given by z l = sin z , as in Example 12 of Section 8.21. Under this mapping, z = -An maps on z, = -1, z = $n on z r = 1, and the whole boundary onto the real-axis of the z l plane. The new problem in the zl plane requires a function u harmonic in the upper half-plane, with boundary values: u = 2 for - 1 < xi < 1 , u = 0 for x l > 1 and for x l < 0 . This is solved as in Example 4 above by (8.103). Hence 2 21-1 u = - arg 7r z1+1

and, in the z plane,

2

sinz-1 n arg sinz 1 '

u=-

+

To apply this formula, the simplest procedure is to use the diagram of Fig. 8.30, showing the mapping from the z plane to the zl plane. For each z , we locate the corresponding zl = sin z , measure the angle

as in Fig. 8.35, and divide by in.The answer can also be written explicitly in real form:

2

= - arctan A

2 cos x sinh y sin2x cosh2 y + cos2x sinh2y - 1 '

EXAMPLE 7 u ( x , y ) harmonic for l z l < l , u = l for r = l , c r < 8 < / 3 as in Fig. 8.38; u = 0 on the rest of the boundary. The function

Chapter 8 Functions of a Complex Variable v

t

Figure 8.38

Example 7.

maps It(< 1 on the half-plane Im (zl) r 0. The point ela has as image zl = m; the point elB has as image zl = 0; the arc u < B < /lcorresponds to the negative real axis in the z, plane. Thus the solution is given by

This rather clumsy formula is the formula for the circle corresponding to (8.103) for the half-plane. By use of geometry, this can be written in the much simpler form:

where s is the arc length of the arc on lzl = 1 determined by the chords through e'" and z and through elB and z ; see Fig. 8.38. This problem could also have been solved directly by the Poisson integral formula (8.101): u(ro, 00) = ix

1 - ro2 dB. I + r i - 210cos(Bo - 8)

/.'

The integration, though awkward, can be carried out with elementary functions.

PROBLEMS 1. Solve the following boundary value problems: a) u harmonic and bounded in the first quadrant, lim u(x,y) = 1 f o r O < x < 1, v+o+

'101

lim u(x,y)=O forx > 1, v+o+

lim u(x, y) = 0 for y > 1.

x-+o+

Sec Fig. 8.39.

609

610

&

Advanced Calculus, Fifth Edition

Figure 8.39

Figure 8.40

b) u harmonic and bounded in the sector: 0 < 8 < in, with boundary values 1 on the x axis and 0 on the line y = x , as in Fig. 8.40. c) u harmonic and bounded for 0 < 8 < 217, with limiting value of 1 as z approaches the positive real axis from the upper half-plane and - 1 as z approaches the positive real axis from the lower half-plane. d) u harmonic and bounded in the strip: 0 < y < 1with boundary values Ofor y = 0,x < 0 andfory = 1 , x Oandfory = 1 , x > 0. e) u harmonic and bounded in the domain: 0 < 8 < rr, r > 1 with boundary values of 1 on the real axis and - 1 on the circle. 2. a) Verify that the transformation

maps the half-plane Im (z) > 0 in a one-to-one conformal manner on the w plane minus t h e l i n e s v = 2 n , u < - 1 a n d v = O , u < -1. b) Find the electrostatic potential U between two condenser plates, which are idealized as two half-planes perpendicular to the uv plane along the lines v = a , u < 0 and v = 0, u < 0, if the potential difference between the plates is Uo; that is, solve the boundary value problem: U(u, a ) = Uo for u < 0 and U(u, 0) = 0 for u < 0, U(u, v) harmonic in the remaining portion of the uv plane. 3. a) Show that the transformation w = z2 maps the domain: Im ( 2 ) > 1 in a one-to-one conformal manner on the parabolic domain

611

Chapter 8 Functions of a Complex Variable b) Let a solid be idealized as an infinite cylinder perpendicular to the uv plane, whose cross section in the uv plane is the region u 5 v2. Let this solid be in temperature equilibrium, with temperatures TI maintained on the part of the boundary surface where v 0 and T2 where v < 0. Find the temperature distribution inside the solid; that is, solve the boundary value problem: T(u, v) harmonic for u c v2 with boundary values T(u, v) = TI for u = v2, v > 0, and T(u, v) = T2 for u = v2, v < 0. 4. a) Verify that the functions (in polar coordinates)

are harmonic for r < 1 and have the boundary values cos no, sin no for r = 1. b) Let h(0) be continuous and have period 2n and let its Fourier series converge to h(0) for all 8 : 1

h(0) = -ao 2 a,, =

00

+ x ( a n cos no +'bn sin no),

'/' IT

n=l

h(0) cos no do,

1

bn = -

-H

/"

h(0) sin no do.

-H

Solve the boundary value problem: u(r, 0) harmonic for r < 1, lim u(r, 0) = h(8) as . ?\a8 r -+ 1. [Hint: Use the result of a) to construct the function

Verify convergence to h(0) with the aid of the remark following Theorem 36 in Section 6.15.1

+

5. a) Verify that the function (8.104) has the limiting value :(hi hj+,) if the point (x, y) approaches the point (xi, 0) along the line x = x j , y > 0. b) What is the limiting value if the approach to (xi, 0) is along a line making an angle a > 0 with the positive x direction? C) Discuss the limiting values of the function (8.107) as (x, y) -+ (rl, 0) at a point of discontinuity tl of h(t). Remark Problem 5 has an analogue in terms of the Poisson integral formula for the circle. As Problem 4 suggests, the topic is closely related to Fourier series. Actually, the properties of the Poisson integral stated in Section 8.22 above imply that, if h(0) is piecewise continuous, ,I: then for the series obtained in Problem 4b),

lim ~ ( r0), = h(OO)

r-t 1

e+eo

Cf

at every point of continuity of h(Q)and

lim ~ ( r00) , =

r-t I

+

h(@0+) Moo-) 2

at each point of discontinuity of h(0). The Fourier series for h(0) may fail to converge to h(Q)for all 9 under these assumptions, but the function h(0) can be recovered from the series by multiplying the terms by r n and letting r + 1. This process is called Poisson (or Abel) summation of the series. (See Volume 1, Chapter 3, Section 6 of the book by Zygmund listed at the end of the chapter.)

0 612

Advanced Calculus, Fifth Edition 6. Show that the Poisson integral formula (8.101)is transformed into the formula (8.107) for the half-plane, if the circle lzl < 1 is mapped on the half-plane Im (zf)> 0 by the equation .l+z Z =I-. .1-z [Hint: Write (8.101) in the form f

a

where z = eie.The change of variable must be made both for zo and for z:

Show that f (zo)becomes

where h

= h[z(t)].]

As pointed out in Problem 2 following Section 5.15, the conditions div V = 0,

curl V = 0

for a two-dimensional vector field V = ui - vj are the same as the Cauchy-Riemann equations for u and v, so that the complex function u + i v = f ( z ) is analytic. These equations therefore describe an incompressible, irrotational, steady 2-dimensional fluid motion with velocity vector V. If we restrict attention to a simply connected domain, then f (z) has an indefinite integral F(z) (determined up to a constant). If we write

F = 4(x, y) + i1CI.(x9Y), then

a@ - i. a4 =u Ff(Z)= ax ay

-

-

+iv.

Hence

or The function 4 is termed the velocity potential and F(z) is termed the complex velocity potential. We can write

Hence the conjugate of the derivative of the complex velocity potential is the velocity vector:

Chapter 8 Functions of a Complex Variable

Figure 8.41

Flow past an obstacle.

The lines @(x, y) = const are called equipotential lines; they are orthogonal to the velocity vector grad Q, at each point. The lines $(x, y) = const are called stream lines; the velocity vector is tangent to such a line at each point, so that these stream lines can be considered as paths of actual fluid particles. Conformal mapping can be applied to hydrodynamics problems in several ways. First of all, particular problems can be formulated as boundary value problems and solved with the aid of conformal mapping as in the preceding section. Second, starting with a known flow pattern, one can obtain a variety of others in an empirical way by simply applying different conformal transformations. We treat here briefly only one example of the first type. We consider the problem of flow past an obstacle, as suggested in Fig. 8.41. The domain D of the flow is the exterior of a (piecewise smooth) simple closed curve C. Since the domain is not simply connected, we cannot be sure that there is a (single-valued) complex potential f (2). It will turn out that, if we make an appropriate assumption about the flow "at co,"then f (z) does exist. The natural assumption is that the flow approaches a uniform flow at constant velocity at m. Hence f (z) = F1(z)is analytic at m:

a-2 F ( z ) = c o n s t + a o z + a - ~ L o g z - -+..., Z

lzl > R .

If f (z) is to be single-valued, the term in Log z must not be present; hence we also assume the flow to be such that a-1 = 0, so that

The constant a0 is precisely the value of f at 00, so that LSo is the velocity of the limiting uniform flow. The stream function p+ = Im [F(z)] is constant along C, since (in the absence of viscosity) the velocity vector must be tangent to C. It would be natural to formulate a boundary value problem for $, but it is simpler to remark that the assumption (8.11 1) about the behavior of f ( z ) at 00 and the condition: Im [F(z)] = const on C imply that zl = F(z) maps D in a one-to-one conformal manner onto a domain Dl of the z , plane. This is established by the argument principle (Problem 3 below). Since Im (zl) = y , is constant on C, the image of C must be a line segment

614

Advanced Calculus, Fifth Edition

Figure 8.42

Slit-domain.

yl = const in the x l y l plane; thus Dl consists of the entire z l plane minus a slit, as in Fig. 8.42. Conversely, if F ( z ) is analytic in D and maps D in one-to-one fashion , F must have a pole of first order at oo and necessarily on such a slit-domain D l then = Im [ F ( z ) ] is constant on C; thus every mapping of D onto a slit-domain Dl provides an appropriate complex velocity potential F ( z ) .

+

EXAMPLE Let C be the circle: lz 1 = 1. Then

maps D onto a slit-domain D l , the slit lying on the real axis (cf. Example 13 of Section 8.21). Hence this is an appropriate potential for the flow past the circle. The stream function is . = aoy - ao- Y x2 + y 2 '

+

the stream lines are shown in Fig. 8.3 1 (Section 8.21). It can be shown that, given C and the velocity iio at oo, a mapping function F ( z ) having the expansion (8.1 1 1 ) at oo exists and is uniquely determined up to an additive constant [which has no effect on the velocity vector F 1 ( z ) l .A similar theorem holds for flow past several obstacles, bounded by curves C I , . . . , C,; the complex velocity potential maps the domain of the flow onto the zl plane minus n slits. For proofs one is referred to the book of Courant listed at the end of the chapter.

8.25 APPLICATIONS OF CONFORMAL MAPPING IN THE THEORY OF ELASTICITY Two-dimensional problems in the theory of elasticity are reducible to solution of the biharmonic equation

+

or, in terms of the operator v2= ( a 2 / a x 2 ) ( a 2 / a y 2 ) ,

615

Chapter 8 Functions of a Complex Variable The function U is Airy's stress function; its second derivatives

give components of the stress tensor, which describes the forces acting on an arbitrary plane cross section of the solid being studied. Solution of (8.112) in a domain D is equivalent to the solution of two equations:

"$

The solutions P of the second equation are harmonic functions. Furthermore, if U , , U2 satisfy the first equation for given P, then

hence the solutions of the first equation are of form

Ul

+ w,

where U 1is a particular solution and W is harmonic. Now let harmonic functions u -* and v be chosen, if possible, so that

Then v2(xu

av + y v ) = x v 2 u + 2-au + y v 2 v + 2- = 4 P ax ay

Hence 1 4

Ul = -(xu

+yv)

is the desired particular solution, provided u and v can be found. If D is simply connected, one can choose Q so that P i Q = F ( z ) is analytic in D; then

+

defines (up to additive constants) harmonic functions u and v such that a u l a x = a v / a y = P and U 1can be written as follows: 1 01= ;(xu

1 + y v ) = -Re [if (z)]. 4

Finally, the solutions U of (8.1 12) can be written thus:

Advanced Calculus, Fifth Edition The factor of

$ can be absorbed in f ( z ) , and we have the conclusion: iff ( z ) and

g(z) are analytic in domain D, then

U

= Re [Z f ( 2 )

+ g(z)l

(8.113)

is biharmonic in D; i f D is simply connected, then all biharmonic functions in D can be represented in this form. The boundary value problems for the stress function U can be formulated in terms of the analytic functions f and g. If D is mapped conformally on a second domain D l , the problem is transformed into a boundary value problem for D l . By mapping onto a simple domain D l , such as the half-plane or unit circle, we reduce the problem to a simpler one. Hence, just as for the Dirichlet problem, conformal mapping is a powerful aid. The importance of (8.113) is that it expresses U in terms of analytic functions, which remain analytic if a conformal change of variables is made. A biharmonic function does not in general remain biharmonic under such a transformation. For further details on the applications to elasticity one is referred to Chapter 5 of the book by Sokolnikoff listed at the end of the chapter.

In general, conformal mapping can be helpful in solving all boundary value problems associated with the Laplace equation or the more general Poisson equation

in the plane. The crucial fact is that harmonic functions remain harmonic under conformal mapping. While the boundary values may be transformed in a complicated manner, this disadvantage is usually offset by the possibility of simplifying the domain by an appropriate mapping. We mention here one example of the possibilities and refer to the books by Kellogg and by Frank and von Mises listed at the end of the chapter for further information. Let it be required to find a function u ( x , y ) harmonic in a given domain D bounded by a smooth simple closed curve C and satisfying the boundary conditions: u has given values h ( x , y) on an arc of C ; aulan = 0 on the rest of C, where n is the normal vector to C . The condition aulan = 0 is equivalent to the condition that the level curves of u are orthogonal to C ; hence this condition is invariant under conformal mapping. To solve the problem, we map D, if possible, on the quarter-plane:xl > 0, yl > 0, so that the arc on which aulan = 0 becomes the yl axis. The function h ( x , y ) becomes a function h ( X I ) , giving the values of u for y, = 0 . We now solve the boundary value problem: u(xl, y I ) harmonic in the upper half-plane; u ( x l ,0 ) = h ( x l) for xl > 0 and u ( x l ,0 ) = h l ( - x l ) for xl < 0 . In other words, we reject the boundary values in the line x , = 0 . The function u obtained is harmonic in the quarter-plane and has the correct boundary values for xl > 0 . Furthermore, (8.107) shows that u ( x l ,y l ) = u ( - x l , y l ) , that is, u shows the same symmetry as the boundary values. This implies that a u / a x l = 0 for xl = 0 , that is, aulan = 0 on the boundary

,

'Chapter 8 Functions of a Complex Variable of the quarter-plane. Accordingly, u satisfies all conditions. If we return to the xy plane, u becomes a function of x and y which solves the given boundary value problem. This can be shown to be the only bounded solution, provided h is piecewise continuous.

PROBLEMS 1. Prove that

f (z)

(

= ao zeIa

+ - + const z:a l)

(areal)

maps D: lzl > 1 in a one-to-one conformal manner on a slit domain Dl (this is the most general such map). Interpret f as a complex velocity potential. 2. Show that the vector

can be interpreted as the velocity of an irrotational, incompressible flow past the obstacle bounded by the circle x 2 + y2 = 1. Find the complex velocity potential and the stream function, and plot several stream lines. 3. Let w = F(z) be analytic in the domain D outside the simple closed curve C and have a pole of first order at oo.Let F(z) be continuous in D plus C and let Im [F(z)] = const on C. Show that F(z) maps D in a one-to-one fashion on a silt-domain. [Hint: Show that the argument principle of Section 8.17 here takes the form: the increase in arg F(z) as z traces C in the negative direction is 2n times ( N o - N,), where No and N , are the numbers of zeros and poles of F(z)in D plus the point z = oo.Show that the increase in arg [F(z) - wo] is 0 for wo not on the image of C. Hence No - N , = 0; but N , = 1, so that No = 1.1 4. Let U ( x , y ) be biharmonic for x 2 + y 2 -= 1 . Show that U can be expanded in a Taylor series in this domain, so that U is analytic in x and y . 5. Find the equilibrium temperature distribution T in the half-strip: 0 < x < 1, y > 0 if the edge x = 0 is maintained at a temperature To, the edge x = 1 is maintained at a temperature T , , while the edge y = 0 is insulated (aT/an = 0).

8.27

GENERAL FORMULAS FOR ONE-TO-ONE MAPPING SCHWARZ-CHRISTOFFEL TRANSFORMATION

The examples above have indicated the importance of conformal mapping for applications. There still remains the problem of exhibiting a reasonably large class of explicit mappings to meet the needs of applications. Here we give several formulas that help in this direction.

Mapping onto injinite strip with slits. Let real constants h l , h2, . . . ,h,,, h,+, be chosen such that for some m, h l < hZ < . - .< h,;

h , > h,+l >

- . . > h,+l.

(8.115)

617

Advanced Calculus, Fifth Edition

+

As extreme cases one can choose m = 1 or m = n 1 . Let points xl, . . . ,x, be chosen on the real axis such that xl < x2 < . . . < x,. A bounded harmonic function v in the half-plane y > 0 , with the following boundary values for y = 0 : v =hl

forx < X I ,

v=h,+l

forx >x,,

is provided by (8.104):

+

here z l = xl Oi, 2. = x2 in the upper half-plane is

+ Oi, . . . . A corresponding f (z) = u + i v = w analytic 8

'Nl

+ hn+l{in - Log (z - zn)] + a,

1

where a is a real constant. When the constants h l , . . . , h,+l satisfy (8.115), this function gives a one-to-one conformal mapping of the half-plane; for a proof, see Problem 5 below. The image domain Dl is bounded by lines v = const.

EXAMPLE 1

z

Log(z+1)+3~~~2+1

I

+ 2(in - Log Z)

maps the upper half-plane on the domain Dl shown in Fig. 8.43.

I

-1 Figure 8.43

Range of example.

*U

1

Chapter 8 Functions of a Complex Variable In general the image domain D l lies between the lines v = h*, v = h**,where h* and h**are the smallest and largest of h l , h2, . . . , h,+l ;the boundary of D l consists of these lines plus rays v = const corresponding to the remaining h's. Every such image domain D l can be obtained by appropriate choice of the numbers xl , . . . , x,, a . For the general mapping theorem (Section 8.22) ensures existence of the map; since v is bounded and has the given boundary values, the function v must have the form given above, so that f must have form (8.116). Mapping onto sector with radial slits. If h** - h* 5 2n, so that the domain D l has width at most 277, then the function ew takes each value at most once in D l . Hence f (z) = exp [f (z)] also gives a one-to-one mapping, provided (8.1 15) holds. The function f (z) can be written as follows:

~ exp ' [kl Log the powers of (z - zl), . . . are taken as principal values: (z - ~ 1 ) = (z - z l ) ] The function ew maps D l onto a domain D2 lying in the sector: h* < arg w2 < h** and bounded by the edges of the sector and other lines and rays as suggested in Fig. 8.44; hence w2 = F(z) maps the upper half-plane on the domain D2. When h**-h* = 21r, the sector has opening 2n and the boundary lines coincide. Schwarz-Christoffel mappings. Another important class of mappings is obtainable from f (z), namely, the mappings

Jzo

or, more explicitly,

& ,%

Figure 8.44

Range o f f (z).

r

h-'

r

-8,

619

620

Advanced Calculus, Fifth Edition where A and B are complex constants. The function w = G(z)defines the SchwarzChristoflel transformation. Every one-to-one conformal mapping of the upper halfplane onto a simply connected domain bounded by straight lines and line segments is representable in the form (8.1 18). This includes every mapping onto the interior of a polygon. The function G(z) satisfies the boundary conditions: arg Gf(z)= h2 for y = 0, xl < x < x2, . . . ,

for = e f ( ~=) eu+iu ,

arg G'(z) = u = Im If (z)l.

/

The quantity arg Gf(z)measures the amount by which directions are rotated in going from the z plane to the w plane (Section 8.20). Hence along the boundary the amount of rotation is piecewise constant; accordingly, each interval x < X I , x l < x < xz, . . . must be mapped on a straight line by w = G(z). The numbers h 1 , h2,. . . give the angles from the direction of the positive real axis to these lines, as suggested in Fig. 8.45 for the case of a convex polygonal domain; the numbers

+ +

+

are successive exterior angles of the polygon; the (n 1)th exterior angle is hl - hn+,. It can happen that hn+]= hl 27r, in which case the polygon has only n vertices; otherwise, for a proper convex polygon hl n < h,+, < h l 217.Since

+

2n

the two cases are as follows:

+ k2 + . - + kn = 2 1 < kl + . - . + kn < 2

kl

Figure 8.45

a

Convex polygonal domain.

(n vertices), (n

+ 1 vertices).

+

Chapter 8 Functions of a Complex Variable

+

The vertices wl, . . . , w, of the polygon are the images of zl, . . . , z,. The (n 1)th vertex, if present, is the image of z = oo. We note that, for a convex polygon, 0 < k , n < rr, 0 < kZrr < rr, . . . , SO that the k's are between 0 and 1. Given any set of k's such that

the corresponding function w = G(z) determines a one-to-one mapping of the upper half-plane onto the interior of a convex polygon with exterior angles k l n , k2n, . . . . Conversely, given a polygon (convex or not) with vertices w ,, . . . , w,+, , there exists a one-to-one mapping w = G(z) of the upper half-plane onto the interior of the polygon. Determination of the points z l , . . . ,z, and the constants A and B is in general a difficult problem in implicit equations; for special cases, as illustrated in the problems below, the constants can all be determined. For more details, see pages 253-258 of the book by Ahlfors listed at the end of the chapter. As mentioned above, the image domain can also be bounded by infinite lines. Such a case is suggested in Fig. 8.46, on which the values of the h's are shown. The boundary of the domain can be regarded as a polygon, some of whose vertices are at m. In the example shown, there are 10 vertices in all. As the example shows, some of the sides can overlap, so that arg G'(z) has values differing by frr on opposite "edges" of the side. Let a simply connected domain D of the w plane be given and let it be required to map the half-plane y > 0 of the z plane in a one-to-one conformal manner on D. It follows from the general mapping theorem of Section 8.22 that such a mapping can be found, provided D is not the finite w plane. In fact, infinitely many such mappings can be provided. Let w = f (z) be one such mapping and let z = g(zl) be a linear fractional transformation [cf. (8.99)] of the half-plane Im (zl) > 0 on the half-plane Im(z) > 0. Then w = f [g(zl)] = h(zl) maps Im(zl) > 0 on D; that is, w = h(z) maps Im (z) > 0 on D. For every choice of the mapping g of half-plane on half-plane, we get a new mapping h of the half-plane on D. It can be shown that all mappings of the half-plane on the given domain D are obtainable from a single mapping f by compounding in this way with a linear fractional transformation g of half-plane on half-plane.

Remark.

Figure 8.46

J

Passible range of G ( z ) .

*

,

621

622

Advanced Calculus, Fifth Edition One can single out a particular mapping f by imposing additional conditions. For example, one can require that three specified points z l , ~ 2 ~ on 2 3the real axis correspond to three specified points wl, 202, wg on the boundary of D, provided the cyclic order of the triples matches the corresponding positive directions on the boundaries. The justification for this rule is given by the fact that one can map the half-plane onto itself to take three given points on the real axis to three given points on the real axis, provided the cyclic orders match (cf. Problem 9 following Section 8.21). The points z,, . . . , z,, oo play special roles in the transformations (8.1 15), (8.117), (8.118). In seeking a transformation of one of these types onto a given domain, one can assume three of the points at convenient positions, provided the cyclic order chosen agrees with the sense-preserving property of the transformation. It will be found most convenient to always make oo a special point, so that h,+l # h 1 and, for (8.1 18), h,,, # h , 2n.The other two special points can be chosen, for example, as 0 and 1.

+

Maps onto circle with teeth. As a final example of a class of explicit one-to-one conformal mappings, we mention functions w = H ( z ) having the following form:

Here the numbers k, and a, are real and k, > 0 , O < a, < 1 for s = 1, . . . , n; the numbers z , , . . . , z, represent distinct points on the circle lzl = 1; the principal value of the a, power is used. The function H ( z ) is then analytic for lzl < 1 and is moreover one-to-one in this domain (Problem 6 below). The image domain is approximately the region I w 1 < 1 plus n sharp "teeth" projecting from it; the points of the teeth are the images of the points z l , . . . , z,. The smaller each a,, the sharper the corresponding tooth. Further properties of these functions and other mapping classes are described in a paper by P. Erdos, F. Herzog, and G. Piranian, Pacific Coast Journal of Mathematics, Vol. 1 (1951), pp. 75-82.

PROBLEMS 1. Verify that the following functions define one-to-one conformal mappings of the upper

half-plane and determine the image domains: a) w = 2Log(z + 1) - Logz b) w =Log+ + 2 ~ o g ~ + 3 L -2Log(z-3) o g ~ e) w =

Jw

(pficipal value).

2. Determine a one-to-one conformal transformation of the half-plane Im (z) > 0 onto each of the following domains: a) the domain bounded by the lines v = 0, v = 2 and the ray v = 1,0 ( u < co.[Hint: Seek a transformation of form (8.116),with z1 = 0, z2 = 1, hl = 0, h2 = 1, h3 = 2.1 b) the domain bounded by the lines v = 0, v = 2 and the rays: v = 1, -co -= u ( - 1; v = 1, 1 5 u < co.[Hint: Use (8.116) with zl = -1, z2 = 0,z3 = p > 0, hl = 0, h2 = 1, h3 = 2, h4 = 1 and determine p so that the mapping is correct.]

Chapter 8 Functions of a Complex Variable C)

623

the first quadrant of the w plane minus the segment from

[Hint: First apply the transformation wl = Log w and then map the half-plane on the resulting domain in the wl plane.] 3. Show that each of the following transformations of the upper half-plane can be regarded as a special case of the Schwarz-Christoffel transformation: 1 Re(w) < ~n c) the transformations of form (8.1 16) d) the transformations of form (8.1 17). 4. With the aid of the Schwarz-Christoffel transformation, determine a one-to-one conformal transformation of the half-plane Im (z) > 0 on the w plane minus the lines v = 2n, u < - 1 and v = 0, u < - 1 [cf. Problem 2a) following Section 8.231. 5. Prove that, if (8.1 15) holds, then (8.1 16) defines a one-to-one conformal transformation of the half-plane Im(z) > 0. [Hint: Prove that each nonconstant term g of f = - C k, Log (z - 2,) const satisfies the inequality: Re [i(z - zm)g'(z)]> 0 in the halfplane. Then apply test VI of Section 8.20.1 6. Let w = u + i v = H(z) be defined by (8.1 19). under the conditions described above. Show that H(z) is analytic and one-to-one for lzl < 1. [Hint: Show that each term h of H(z) satisfies the condition: Re [hl(z)] > 0 for lzl < 1. Now apply test V of Section 8.20.1

b) w = sin-' z, -in

i

+

Suggested References Ahlfors, Lars V., Complex Analysis, 3rd ed. New York: McGraw-Hill, 1979. Betz, Albert, Konfor-me Abbildung, 2nd ed. Berlin: Springer, 1964. Bieberbach, Ludwig, Einfuhrung in die konforme Abbildung. Berlin: Walter de Gruyter, 1967. Bieberbach, Ludwig, Lehrbuch der Funktionentheorie (2 vols.), 4th ed. New York: Chelsea, 1945. Churchill, Rue1 V., and James W. Brown, Complex Variables and Applications, 5th ed. New York: McGraw-Hill, 1990. Courant, R., Dirichlet's Principle, Conformal Mapping and Minimal Surfaces. New York: Springer, 1977. Frank, P., and R. von Mises, Die DifSerentialgleichungen und Integralgleichungen der Mechanik und Physik, 2nd ed. New York: Dover, 1961. ,~ in MathematicalAnalysis, Vol. 11, Part 1 (transl. by E. R. Hedrick Goursat, ~ d o u a r dCourse and 0. Dunkel). New York: Dover, 1970. Heins, Maurice, Selected Topics in the Classical Theory of Functions of a Complex Variable. New York: Holt, Rinehart and Winston, 1962. Hurwitz, A. and R. Courant, Funktionentheorie, 4th ed. Berlin: Springer, 1964. Kaplan, Wilfred, "Close-to-convex Schlicht functions." Michigan Mathematical Journal, Vol. 1, pp. 169-185, 1952. Kellogg, 0. D., Foundations of Potential Theory. New York: Springer, 1967. Knopp, Konrad, Theory of Functions ( 2 vols., transl. by F. Bagemihl). New York: Dover, 1953. Nehari, Zeev. Conformal Mapping. New York: McGraw-Hill, 1952. Nevanlinna, Rolf, Analytic Functions, transl. by P. Emig. New York: Springer, 1970. Osgood, W. F. Lehruch der Funktionentheorie (2 vols.). New York: Chelsea, 1965.

,

dig

624

Advanced Calculus, Fifth Edition Picard, p mile, Traitk dPAnalyse,3rd ed., Vol. 11. Paris: Gauthier-Villars. 1922. Sokolnikoff, I. S., Mathematical Theory of Elasticity, 2nd ed. New York: McGraw-Hill, 1956. Titchmarsh, E. C., Theory of Functions, 2nd ed. Oxford: Oxford University Press, 1953. Whittaker, E. T. and G. N. Watson, A Course of Modern Analysis, 4th ed. Cambridge: Cambridge University Press, 1950. Zygmund, A., Trigonometric Series, 2 vols. London: Cambridge University Press, 1959.

Ordinary Differential Equations

9.1 DIFFERENTIAL EQUATIONS * '+

An ordinary differential equation of order n is an equation of form

expressing a relation between x, an unspecified function y(x), and its derivatives y', y", . . . through the nth order. For example,

4

are ordinary differential equations of orders 2 and 3, respectively. In order for the differential equation (9.1) to have significance, it is necessary that the function F be defined in some domain of the space of the variables on which it depends. In this chapter we shall consider only equations that can be solved for the highest derivatives and written in the form: y(n)

= F(x, y , y ' , . . . , y'n-").

(9.4)

Equation (9.2) is at once reducible to this form. Equation (9.3) is a quadratic equation in y"'; by solving this quadratic equation we obtain two different equations for y"', that is, two equations of form (9.4). We say that (9.3) is a differential equation of degree 2, whereas (9.2) is of degree 1. Systems of differential equations will also be considered (Section 9.5). 625

626

Advanced Calculus, Fifth Edition The term ordinary is used here to emphasize that no partial derivatives appear since there is just one independent variable. An equation such as

would be called a partial diflerential equation. Chapter 10 is devoted to such equations. The applications of ordinary differential equations to physical problems are numerous. The equations of dynamics are relations between coordinates, velocities, accelerations, and time and hence give differential equations of second order or systems of higher order. Electric circuits obey laws described by differential equations relating currents and their time derivatives. Servomechanisms, or control systems, are combinations of mechanical and electrical (and perhaps other) components and can be described by differential equations. Problems involving continuous mediafluid dynamics, elasticity, heat conduction, and the like-lead to partial differential equations.

By a solution or particular solution of (9.1) is meant a function y = f ( x ) , a < x < b, having derivatives up to the nth order throughout the interval and such that (9.1) becomes an identity when y and its derivatives are replaced by f ( x ) and its derivatives. Thus y = ex is a particular solution of (9.2), and y = x is a particular solution of (9.3). For most of the differential equations to be considered here it will be found that all particular solutions can be included in one formula: where el, . . . , en are "arbitrary" constants. Thus for each special assignment of values to the c's, (9.6) gives a solution of (9. I), and all solutions can be so obtained. (The range of the c's and of x may have to be restricted in some cases to avoid imaginary expressions or other degeneracies.) For example, all solutions of (9.2) are given by the formula y = cle-x

+ c2e-2X+ ex;

(9.7)

the solution y = ex is obtained when c , = 0, c2 = 0. When a formula such as (9.6) is obtained, providing all solutions, it is called the general solution of (9.1). For the equations considered here it will be found that the number of arbitrary constants equals the order n. The presence of arbitrary constants should not be surprising, for they occur in the simplest differential equation:

All solutions of (9.8) are obtained by integration:

Chapter 9 Ordinary Differential Equations Here there is one arbitrary constant: cl = C. This can be generalized to equations of higher order, as the following example shows:

y" = 2ox3.

(9.10)

Since y" is the derivative of y', one concludes by integrating twice in succession that

y1=5x4 +el,

y = x5 +clx+cz.

(9.11)

9.3 THEBASICPROBLEMS rn EXISTENCE THEOREM For a given differential equation or a given system of equations the fundamental problem is that of finding all solutions. This is a formidable task which can be satisfactorily completed only for a very few simple differential equations. Moreover, the meaning of the phrase "finding all solutions" is not as obvious as it might seem at first. For with the aid of a digital computer, one can compute to high accuracy as many desired solutions as needed for a particular practical problem. Besides the general problem of finding all solutions, there are two special problems of great importance: The initial value problem and the boundary value problem. For the Eq. ( 9 . 1 ) the initial value problem is as follows: Given a value xo of x and n constants: (n-1)

YO?Y;,...,YO

9

a solution y = f ( x ) of ( 9 . 1 ) is sought such that y = f ( x ) is defined in an interval Ix -xol < S(S > 0)and = yr-l) (9.12) f ( x o ) = y o , f l ( x o ) = y ; , ... f 7

It x is a variable representing time and xo = 0, then (9.12) imposes n conditions on the solution at time 0 or "initial conditions." THEOREM 1 Let an ordinary differential equation of order n be given in the form

Y'") = ~ ( xy , y', . . . , y'"-")

(9.13)

and let the function F be defined and have continuous first partial derivatives in a domain D of the space of its variables. Let ( x 0 . yo, yh, . . . , Y(II-l))be a point of D. Then there exists a function y = f (x), xo - 6 < x < xo 6 (6 > 0 ) , which is a particular solution of ( 9 . 1 3 ) and satisfies the initial conditions (9.12). Furthermore, the solution is unique; that is, if y = g ( x ) is a second solution of (9.13) satisfying (9.12),then f ( x ) = g ( x ) wherever both functions are defined.

+

For a proof of this theorem, see Section 9 . 9 . When the general solution is explicitly known, the solution of the initial value problem is relatively simple. For example, from the general solution ( 9 . 7 ) of (9.2), one obtains a particular solution such that y = 1 and y' = 2 for x = 0 by solving the equations

+

Accordingly, cl = 1 , cz = - 1, and y = e-" - e-2" e' is the particular solution sought. In general, Theorem 1 gives a way of testing a formula that is believed to

627

Advanced Calculus, Fifth Edition give the general solution (all solutions): Zftheformula dclfines solutions adprovides one solution for each allowable set of initial conditions, then it is indeed the general solution. It should be noted that the initial conditions (9.12) impose n conditions on the function f (x) for a chosen value xo of x. Since the general solution (when it can be found) has n arbitrary constants, one is always led to n simultaneous equations in n unknowns, which "in general" have one and only one solution. One can impose some of the n conditions at one value, xo,of x and the others at a second value, X I . Then one seeks a particular solution y = f ( x ) defined for xo 5 x 5 xl and satisfying the given conditions at xo and x l . This is precisely the boundary value problem for (9.13). For example, let the problem be to find a solution y = f (x) of (9.2) that satisfies the two conditions f (0) = 0,f (1) = 0. One is immediately led to the two equations 0 = c l e P 1 c2e-2 e ; 0 = cl c2 1 , these have the one solution: c l = -1 - e - e2, c2 = e e2;SO that y = (- 1 - e - e2)e-" (e e2)e-*" ex is the solution sought. The conditions under which the general boundary value problem has a unique solution are quite involved and will not be considered here.

+

+ +

+ +

+

+

+

Elementary methods for the jirst order equation. It is assumed that the reader has studied the special methods for obtaining the general solution of a first order equation. Some of these are reviewed in Problems 6 to 9.

PROBLEMS 1. Find the general solution of each of the following differential equations: d a) = eZX- x b) d, x2,Y= O c) =x

3

2

d)

d" y

==o

e) = d nY= l

f)

$=xI

2. Find a particular solution satisfying the given initial conditions for each of the following differential equations:

2

a) d = sinx, y = 1 forx =O;

d

y=lforx=O. 3. Find a particular solution of the differential equation satisfying the given boundary conditions: c) $ = y ,

4. Verify that the following are particular solutions of the differential equations given: a) y = sinx,foryU+y =O; b) y = e2x,for y" - 4y = 0; c) y = cl cosx c2 sinx (cl and c2 any constants), for y" y = 0; , d) y = cle2x c2e-" for y" - 4y = 0.

+

+

+

629

Chapter 9 Ordinary Differential Equations

5. State the order and degree of each of the following differential equations:

6. (Separation of variables) If F ( x , y ) can be written as - P ( x ) / Q ( y ) , then y' = F ( x , y) can be replaced by P ( x )d x Q ( y )d y = 0, and the general solution can be obtained as !P(X)dx +/ ~ ( ~ ) = d const. y Apply this procedure to the following equations: b ) y l = sinxcosy a) y' = eX+Y d ) y' = Y-2 C ) Y' = (Y - I)(Y - 2 )

+

7. (Homogeneous equations) I f y' = F ( x , y ) = G ( v ) ,where v = y / x , then the substitution y = x v (leading to y' = xu' v ) leads to the equation xu' v = G ( v ) ,which can be solved by separating variables. Apply this procedure to the following equations:

+

+

a) y1 =

b ) xy' - y = xeyfX c) ( 3 y~ 3)d x ~ (x3 ~ 3 x y 2 )d y = 0 8. (Exact equations) A first order equation, written in the form P ( x , y ) d x Q ( x , y ) d y = 0, is said to be exact if P ( x , y ) d x Q ( x , y ) d y = du for some function u ( x , y), so that aulax = P, au/ay = Q. The general solution is then given in implicit form as u ( x , y ) = c. One can test for exactness by seeing whether a P l a y = a Q l a x . If this condition holds in a simply connected domain D, then as in Section 5.6, u is given by the equation

+

+ +

+

+

2

on an arbitrary path in D. Verify that the following equations are exact and find all solutions: a ) 2xy d x ( x 2 1 ) d y = 0; b) ( 2 x + y ) d x + ( x - 2 y ) d y = O ; c) [ x c o s ( x + y ) + s i n ( x + y ) ] d x + x c o s ( x + y ) d y =O. 9. (First order linear equation) This is the equation y' p(x)y = q ( x ) .After multiplying by v ( x ) = exp p ( x ) d x it can be written as (v(x)y)' = q ( x ) v ( x ) ,so that the solutions are given by v ( x ) y = j" q ( x ) v ( x )d x c. Find the general solution of each of the following differential equations: 1 b ) (sin2x - y ) d x - tanx d y = 0 a) dy = sinx

+ +

+

+

+

~)(~'~l)dx+(~~-y+2x)dy=O

d ) % + x = e 2 f

An ordinary linear differential equation of order n is a differential equation of form

It will b e assumed here that the coefficients ao(x),a l ( x ) ,. . . , a,(x) and the righthand member Q ( x ) are defined and continuous in an interval a _< x I b of the x-axis and that ao(x)# 0 in this interval.

630

Advanced Calculus, Fifth Edition The following are examples of linear equations:

+

(a) y" y = sin2x, (b) x~~~~~ - xy' exy,= logx (c)

d'y

+

d'y

-x2

+x32

(x > O),

=o,

(dl y"+ y = 0, (e)

2 +xzy = e'.

If Q(x) = 0, the Eq. (9.14) is said to be homogeneous (with respect to y and its derivatives). Thus (c) and (d) are homogeneous: the other examples are nonhomogeneous. If Q(x) is replaced by 0 in a general equation (9.14), a new homogeneous equation is obtained, termed the homogeneous equation corresponding to (or related to) the given diferential equation. If the coefficients a&), al(x), . . . , a n @ )are all constants, hence independent of x, the Eq. (9.14) is said to have constant coeflcients, even though Q(x) depends on x. Thus (a) and (d) have constant coefficients; the other examples do not. It will be convenient to write the differential equation (9.14) in "operational"

form:

I

d"

7+ . . . + a,-~(x) d x + an(x) [y] = Q(x) or, with the abbreviation:

simply as follows:

+

For example, the equation: xy" 2xy' - 3y = 5 would be abbreviated: L[y] = 5, L = x(d2/dx2) 2x(d/dx) - 3. From the basic rules of differentiation we conclude that L is a linear operator (Section 3.6); that is,

+

where yl(x) and y2(x) are functions having derivatives through the nth order for a 5 x 5 b and e l , c2 are constants. From this we conclude: Ifyl(x) and y2(x) are solutions of the homogeneous equation L[y] = 0, then so also is c l yl(x) c2yZ(x). Hence from known solutions y (x), . . . , y,(x) of the homogeneous equation we can . . c, y,(x) containing n arbitrary constants. construct solutions y = c l y, (x) This has the appearance of a general solution. However, it might be the case that yn(x), for example, is a linear combination of yl(x), . . . , yn-l(x):

+

+ +

where kl , . . . , kn-l are constants. The hypothetical general solution would then really involve only n - 1 constants:

Chapter 9 Ordinary Differential Equations and could not be expected to be the general solution. In order to rule this out we assume that the functions yl ( x ) ,. . . , y,(x) are linearly independent (Sections 1.17, 7.1O), that is, that no one of the functions is expressible as a linear combination of the others or, equivalently, that an identity c ~ y ~ ( x ) + . ~ . + c , y , ( x ) = O ,a i x s b , can hold only if the constants e l , . . . , c, are all 0. If this is the case, then y = cl y l ( x ) . . - c,yn(x) is indeed the general solution:

+ +

THEOREM 2 There exist n linearly independent solutions of the homogeneous differential equation L [ y J= 0 in the given interval a 5 x 5 b. If y l ( x ) ,. . . , yn(x) are linearly independent solutions of the equation L [ y ] = 0 for a 5 x 5 b, then y = cl y l ( x ) . . c, y,(x) is the general solution.

+ +

For a proof, see Section 9.9. The general solution of the nonhomogeneous equation L [ y ] = Q ( x ) can be constructed from the general solution cl y l ( x ) . - . c,y,(x) of L [ y ]= 0 and one particular solution y*(x)of L [ y ] = Q, namely, as the expression

+ +

+

+ + c,yn(x).

Y = y*(x) ~ I Y I ( ~ .) .

(9.17)

For, by linearity, L[yl = Lty*

+ cly, + ... + cnynl = L[y*l+ cl L[y11+ . .. + cnL[ynl +

= L[y*l 0 = Q(x);

furthermore, if y ( x ) is any solution of L [ y ]= Q, then L [ y - y*] = L [ y ]- L[y*]= Q - Q = 0. Hence y - y* is a solution of the homogeneous equation, and y has form (9.17). THEOREM 3 There exists a solution of the nonhomogeneous equation L [ y ] = Q ( x )in the given interval a 5 x 5 b. If y*(x) is one such solution and cly l ( x ) c,y,(x) is the general solution of L [ y ] = 0, then y = y*(x) c l y l ( x ) - c,y,(x) is the general solution of L [ y ]= Q ( x )for a 5 x 5 b.

+ +

+

+ +

For a proof of existence of a particular solution y*(x) we again refer to Section 9.9. In general, existence of solutions is guaranteed by Theorem 1 of Section 9.1; however, one must also show that each solution is defined throughout the given interval a 5 x 5 b. Linear equations with constant coeficients. We assume familiarity with the methods for finding the general solution for these equations. We summarize the procedures. Let the equation be

where ao, . . . , a, are constants, a0 # 0, and Q ( x )is continuous for a! 5 x 5 p. Then one forms the characteristic equation:

+ +

aOrn . - . a, = 0.

'c;*

(9.19)

631

632

Advanced Calculus, Fifth Edition

3

After finding the n roots of this equhtion, one assign9

,

.'

..

! I

( I ) to each simple real root r the function erx; (11) to each pair a fbi of simple complex roots the functions eaxcos bx, eaxsin b x ; (111) to each real root r of multiplicity k the functions e r x ,xerx,. . . , xk-'erx; ( I V ) to each pair a fbi of complex roots of multiplicity k the functions eaxcos bx, eaxsin bx, xeaxcos bx, xeaxsin b x , . . . , xk-' eaxcos bx, xk-'eax sin bx. If the n functions y ( x ) ,. . . , yn(x)thus obtained are multiplied by arbitrary constants and added, then one obtains the general solution of the related homogeneous equation

L [ Y ]= 0

(9.20)

in the form

+ +

Y =c~Yl(x) . C~Y,I(X). (9.21) From this expression, one obtains the general solution of the nonhomogeneous equation (9.18)as

where y*(x)is a particular solution of (9.18).This solution can be found by variation of parameters: where v ; ( x ) ,. . . , vL(x)are chosen to satisfy the linear equations

and v l ( x ) ,. . . , vn(x)are obtained from v ; ( x ) ,. . . , vL(x)by integration.

Remarks. The determinant of coefficients in (9.23) is

This is called the Wronskian determinant of the n solutions y l ( x ) ,. . . , yn(x) of the homogeneous equation. One can show that W ( x ) # 0 for cr i x 5 /? (see Problem 17). The method of variation of parameters can also be used for equations with variable coefficients. Once the general solution of L [ y ] = 0 is known in the form

Chapter 9 Ordinary Differential Equations

(9.21), one can proceed exactly as before to obtain y*(x) and the general solution (9.22). The general solution of L [y] = 0 is often called the complementaryfunction.

PROBLEMS

+ c ~ x - ~x ,> 0, satisfies the differential equation

1. Verify that the function y =

+

x ~ ~4xy'"

+ 2y = 0

for every choice of cl and c2 and that cl and c2 can be chosen uniquely so that y satisfies the initial conditions y = yo and y' = yb for x = xo (no > 0). Hence y = clx-2 + c ~ x - ~ is the general solution for x > 0. 2. Verify that the function y = x2

+ e-x(cl cos 2.x + c2 sin 2x)

satisfies the differential equation y"+2y'+5y = 5 x 2 + 4 x + 2 for all x and that cl and c2 can be chosen uniquely so that y satisfies the initial conditions y = yo, y' = y; for x = xo. Hence the given expression is the general solution.

3. Show that the functions ex, e2', e3Xare linearly independent for all x. [Hint: If an identity

+

+

clex c2e2' c3e3' = 0 holds, then differentiate twice. This gives three equations for cl , c2, c3 whose only solution is cl = 0, c2 = 0, c3 = 0.1 4. Show that the following functions are linearly independent for all x (cf. Problem 3): b) sinx, cosx, sin2x a) x , x 2 , x 3 d) ex, e-x c) ex,xeX,sinh x 5. Determine which of the following sets of functions are linearly independent for all x: a) sinhx, ex, e-' b) cos 2x, cos2 x , sin2x . ., d) x 2 - x + 1 , x 2 - 1 , 3 x 2 - X - 1 c) 1 x, 1 2x, x2 6. Find the general solution: b) y" 4y' = 0 a) y" - 4y = 0 . ! . c) y"' - 3y" + 3y1 - y = 0 d) ~ + Y = O hd-

+

d A

+

+

t

-

f ) d5x+ 3 x = O g) $ + 9 + 7 x = 0

7. Find the particular solution of the differential equation satispiptthe given initial conditions for each of the following cases: a) y l l + y = O , y = l a n d y l = O f o r x = O !I ;i b) $ $ + $ - 3 x = O ,

x=Oand~=lfort=O

8. Find a particular solution of the equation satisfying the given boundary conditions for each of the following cases: a) yll - y f - 6y = 0, y = 1 for x = 0, y = 0 for x = 1 b) y l l + y = ~ , y = l f o r x = O , y = 2 f o r x = c) y u + y = O , y = O f o r x = O , y = O f o r x = n .,-. . . - 6

633

634

Advanced Calculus, Fifth Edition

3

9. Find the general solution by the method of variation of parameters: a) y" - y = ex b) y"' - 6y" + 1 1 y' - 6 y = e" C) yll y = cotx d ) y"+4y = s e c 2 x e) y" - y = logx

+

+

10. Solve the first order linear equation: y' p(x)y = q ( x ) by solving the corresponding homogeneous equation first and then obtaining the general solution by variation of parameters (see Problem 9 following Section 9.3). 11. Verify that y = c l x c2x2is the general solution of the equation

+

and find the general solution of the equation

12. Verify that y = cleX + c2x-I is the general solution of the homogeneous equation corresponding to

and find the general solution. 13. (Operational method) We write D = d l d x , so that the operator L can be written: L = aoDn + . . . + a,-1 D + a,, that is, as a polynomial in the operator D. The sum and product of two operators L 1 , L2 are defined by the equations ( L I L2)[y] = L l [ y ] L 2 [ y ] (, L I L 2 ) [ y ]= L I ( L 2 [ y ] )I .f the coeflcients are constant, then these operations can be carried out as for ordinary polynomials. For example,

+

+

The general rule can be established by induction. To find a solution y * ( x ) of the nonhomogeneous equation

we write the equation in operator form: We then set ( D - 2 ) [ y ]= u , so that ( 2 0 + l ) [ u ]= Q(x). Accordingly, by Problem 9 following Section 9.3,

y = ek

1

e-2xu d x ,

1 I u = -e-jX 2

1

eixQ(x)dx.

Obtain the general solutions of the equations given: a) D(D2 - 9 ) [ y ]= 0 b) ( D 5 - D ~ ) [=~0] C) ( D 2 - 9 ) [ y ]= 8eX d ) ( D - 1)3[y] = 1 e) Problem 9(a) f ) Problem 9(b) The use of operators has been developed into a highly efficient technique. For more information, refer to Chapter 4 of ODE.' " h e book by this author (Kaplan) listed at the end of this chapter will be referred to as ODE.

Chapter 9 Ordinary Differential Equations

14. (Method of undetermined coeficients) To obtain a solution y*(x) of the equation (0' l)[yl = ex, (a) we can multiply both sides by D - 1: (D - 1 ) ( ~ l)[y] ~ = (D - l)[ex] = 0. (b) The right-hand member is 0 because ex satisfies the homogeneous equation (D - l)[v] = 0. From (b) wc obtain the characteristic equation (r - l)(r2 + 1) = 0. Accordingly, every solution of (b) has the form y = cl cosx + c2sinx + c3ex. If we substitute this expression in (a), the first two terms give 0, since they form the complementary function of (a). One obtains an equation for the "undetermined coefficient" cg: 2c3ex = ex, c3 = Accordingly, y* = ;ex is the particular solution sought. The method depends upon finding an operator that "annihilates" the right-hand member Q ( x ) Such an operator can always be found if Q is a solution of a homogeneous equation with constant coefficients: the operator L such that L[Q] = 0 is the operator sought. If Q is of form (po plx . . . pkxk-')eaX(~ cos bx B sin bx), the annihilator is L = ((D - a12 b2)k.For a sum of such terms, one can multiply the corresponding operators. Use the method described to find particular solutions of the following equations: a) y" y = sinx (annihilator is D~ 1) " b) y" + y = e2X(annihilator is D - 2) C) y" y' - y = x2 (annihilator is D ~ ) d) y " + 2 y 1 + y = s i n 2 x + e X .‘ e) y" + 4y = 25xeX f ) y"' - 2y' - 4y = e-X sinx

+

+

i.

+ +

+ +

+

+

+

+

I Cf

15. Prove by variation of parameters or by the method of Problem 14 that a particular solution of the equation

is given by X

B sin (wt - a )

= J0c2

- 02)2+ 4w2h2'

tana=

20h )c2 - ,2 '

O
16. (Application of Laplace transforms) (See Sections 6.25 and 7.19.) For the equation soy(") . . . a, y = g(x) with constant coefficients, one can find the solution for x 2 0, with given initial values yo, yo, . . . , yr-l', with the aid of Laplace transforms. By (7.101).

+ +

+

+ +

Y[y'k'] = S ~ Y [-~[yf-l' ] syf-2' . . . sk-lye] for k = 1, . . . , n. Thus if y(x) is the solution of y" - y = g(x) with y(0) = 1 and yl(0) = 3, then Y [ y " - y] = 2 [ y " ] - Y [ y ] = 2 [ g ( x ) ] = G(s), so that S ~ Y [-~ ] (3 S) - Y [y] = G(s) and 49 [y] = (3 s)(s2 - I)-' c(s)(s2 - I)-'. Now, by partial fractions, (3 s)(s2 - 1)-I = 2(s - 1)-I - (s I)-' = Y [ 2 e X - e-X]. For given g(x) we may be able to find explicitly $(x) such that Y [ 4 ] = G(s)(s2 - 1)-l. Then Y [y] = Y [2eX - eTX $(x)l, so that y = 2eX - e-x #(x). In any case, (s2 - 1)- 1 -- ( 12 ) ( ~ 1)-I - (;)(s I)-' = 9 [ ( i ) e x - (i)e-X] = 9 [ W ( x ) ] and hence . G(s)(s2 - I)-'.= 3[W(x)]3[g(x)] = 2 [ W * g ] by (7.104). Hence 4(x) = W * g =. ,

+

+

+

+

+

+

+

+

635

636

Advanced Calculus, Fifth Edition \

1;

W(u)g(x- u)du. The method can be applied generally to equations with constant ' coefficients, provided that G ( s )exists. Partial fraction expansions lead one to terms of form a(s - b)-k = 9 [ a x k - l e b x / ( k- I ) ! ] . Apply the method described to obtain the solution with given initial values for x > 0, for the following equations: a) y " - 4 y = e 3 x , y = 1 , y l = - l f o r x = O ; .id rdi b) yu - 2y' - 3y = X , y = 0, yt = 0 for x = 0; c ) y" - 4y = g(x), y = 0, y' = 0 for x = 0; d ) y" y = g(x), y = 0, y' = 0 for x = 0. 17. (The Wronskian) Let L be as in Eq. (9.15) and let yl(x), . . . , y,(x) be solutions.of L [ y ] = 0 for a 5 x 5 p. Let W ( x )be as in Eq. (9.24). a ) Show that if yl (x), . . . , yn(x) are linearly dependent, then W ( x )= 0 for a 5 x 5 B. b) Show that if W ( x o )= 0 for one xo on the interval, then yl(x), . . . , y,(x) are linearly dependent on the interval. [Hint: If the n functions are linearly independent, then y = cl y l ( x ) . . . cnyn(x)is the general solution of L [ y ] = 0 (Theorem 2), and hence cl, . . . , cn can be chosen so that y(x) satisfies arbitrary initial conditions for x = xo. From the fact that these linear equations can always be solved, conclude that W(xo)could not be 0.1 C ) Prove that W ( x )satisfies the first order differential equation a0 W' a1 W = 0 on the interval. Hence conclude that if W ( x o )= 0, then W ( x )r 0 on the interval.

+

+ +

+

We consider a system of first order differential equations, of form

Here X I ( t ) , . . . ,x n ( t )are unknown functions o f t , and this set of functions is called a solution of the system (9.25) if all functions are defined over an interval cr < t < B and satisfy (9.25) identically over this interval. For example,

is a system of form (9.25) (with x = X I , y = X Z ) ,and x=2e4',

y=3e

41

, -oo
is a solution, since for all t,

ge4' = 2e4t + 6e4t,

12e4t = 6e4r

+ 6e4r.

+

l ,r

The general system (9.25) is said to have order n. Under appropriate continuity conditions (see Theorem 4 which follows) this system has a unique solution satisfying n initial conditions:

xl (to)= x:, . . . , X , (to) = xn0 . Here to is a selected value o f t , and xp,

. . . ,x:

(9.26)

are given initial values of x l , . . . ,x,.

637

Chapter 9 Ordinary Differential Equations A great variety of systems of ordinary differential equations can be written in the form (9.25). In particular, the single equation of order n (with y' = dy l d t , y" = d2y/dt2,. . .) y(n)

= ~ ( yy',,

. . . , y'n-",

t)

can be replaced by an equivalent system (9.25). We write

so that

dxn dt

- = F(x1, . . . ,x,, t). The pair of equations

fi

can similarly be replaced by the system

of form (9.25). THEOREM 4 In the system (9.25), let the functions f,(xl, . . . , x,, t) (i = 1, . . . ,n) all be defined in a domain D of the (n 1)-dimensional space of the variables X I ,. . . , x,, t. Let these functions be continuous in D and have continuous first partial derivatives af,/axj (i = 1, . . . , n; j = 1, . . . , n) in D. Let (x:, . . . , x:, to) be in D. Then there exists a solution X I = xl(t), . . . , x, = x,(t), to - 6 < t < to S (6 > O), of Eqs. (9.25) satisfying the initial conditions (9.26). The solution is unique: If X I = x;(t), . . . , X, = xi([) is a second solution for to - 6 < t < to 6 satisfying (9.26), then x;(t) xf (t) for to - 6 < t < to 6.

+

+

-

+

+

For a proof, see Section 9.9. An important special case of Eqs. (9.25) is the system of linear equations dx; (9.30) p,(t). dt The existence theorem, Theorem 4, applies if all a;, (t) and p, (t) are continuous for a < t < b; the domain D is then given by

+ +

- = a ; ~ ( t ) x ~... a;,(t)x,

--m<xl<m,

+

..., - m < x , < m ,

a
n

One can show further that for a linear system (9.30), each solution can be chosen to be defined over the whole interval a < t < b. Often, in the linear case, one replaces the open interval a < t < b by a closed interval a 5 t 5 b, and a similar statement applies to this case (see Section 9.9).

638

Advanced Calculus, Fifth Edition The linear system (9.30) can be written in vector-matrix form:

Here A(t) is the n x n matrix (aij@)),while x and p are n x 1 column vectors whose components are functions of t . We assume the stated continuity conditions for a < t < b and seek vector solutions x(t). A set of vector functions x l ( t ) ,. . . , xk(t), a < t < b, is called linearly independent if the condition for scalars cl, . . . , c k , implies cl = 0 , . . . , ck = 0. Otherwise, the set of vector functions is called linearly dependent. We let x l ( t ) ,. . . , x,(t) be n solutions of the homogeneous equation

and form the matrix X ( t ) whose columns are x l ( t ) ,. . . , x,(t), respectively. If x l ( t ) ,. . . , x,(t) are linearly independent, then X ( t ) is nonsingular for each t ; that is, we shall show that

W ( t )= det X ( t ) # 0,

a < t < b.

(9.32)

Suppose W(t0)= 0 , a c to < b. Then we can choose c l , . . . , c, so that and not all c; are 0. For these are n homogeneous linear equations for c1, . . . , c, whose determinant of coefficients is 0. With these ci, form

Then x ( t ) is a solution of the homogeneous equation (9.31). For

Furthermore, by (9.33), x(to) = 0. By Theorem 4 and the succeeding comments, there is a unique solution of Eq. (9.31) for a < t .:b with x(to) = 0. But x r 0 is such a solution. Therefore x(t), as defined by Eq. (9.34), must be identically 0. Hence xl ( t ) ,. . . , x,(t) are linearly dependent, contrary to hypothesis.Therefore X ( t ) is nonsingular, and W ( t )# 0. We now conclude that Eq. (9.34) defines the general solution of Eq. (9.31). For if x = u ( t ) is a solution (for a < t < b), then we select to,a < to < b and choose cl , . . . , c, so that

+ +

Since W ( t o )# 0, we can solve uniquely for c l , . . . , c,,.Thus u ( t )and cl X I ( t ) . c,x,(t) are two solutions of Eq. (9.31) that agree at to. Again by uniqueness they agree for all t in the interval, so that u ( t ) = c l x l ( t )+ . c,x,(t) for a < t < b. Thus Eq. (9.34) does give all solutions of the homogeneous equation.

+

Chapter 9 Ordinary Differential Equations We can now reason as usual that if x*(t) is a solution of the given nonhomogeneous equation (9.30')and x l ( t ) , . . ,x,(t) are linearly independent solutions of the related homogeneous equation (9.3I), then

.

x = x*(t)

+ c l x l ( t )+ - . + c,x,(t)

(9.35)

is the general solution of Eq. (9.30'). The proof is left as an exercise (Problem 8 following Section 9.6). It remains to show that solutions x*(t),x l ( t ) , . . . , x,(t) as described do exist. For the homogeneous equation we select to and then choose x l ( t ) ,x2(t), . . . as solutions such that x l ( t o )= e l = col(1,0, . . . , O), x2(tO)= e2 = col(0, 1,O, . . . , 0), . . . . For the corresponding matrix X ( t ) we have X(to) = I, so that W ( t o ) # 0 . It follows as before that x l ( t ) , . . . , x n ( t ) are linearly independent solutions, as required. For the nonhomogeneous equation, Theorem 4 and the ensuing comments allow us to choose a solution with particular initial value x0 = col (xp, . . . , x:) at a chosen to in the interval a < t < b. This solution is then a desired x*(t). One can also obtain x*(t) from x l ( t ) , . . . , x,(t) by variation of parameters. We seek u ( t ) = col ( u l ( t ) ,. . . , u,(t)) so that is a solution of Eq. (9.30'). If we substitute in Eq. (9.30') and use the fact that

we find

Accordingly,

du

X ( t ) col (u',(t),. . . , uL(t)) = X ( t ) - = p(t). dt Hence since X ( t ) is nonsingular,

~

and u l ( t ) ,. . . , u,(t) can be obtained by choosing indefinite integrals of the components of the right-hand side. We can also select to, a < to < b, and then write

(Sections 4.1,4.5). Thus finally we obtain the variation-ofparametersformula

for the particular solution sought. It is the solution such that x*(to) = 0. Remark.

The matrix X ( t ) can be shown to satisfy the differential equation

639

Advanced Calculus, Fifth Edition We call a nonsingular X(t) formed as before-afundamental matrixfunction for the differential equation (9.30'). (See Problem 5 following Section 9.6.)

WITH CONSTANT COEFFICIENTS 9.6 LINEARSYSTEMS We consider systems such as the following:

Here a l , . . . , c3are constants. The general case would concern n variables XI, that are functions of t : dx;

+ +

. . . ,xn

+

. . - ainx, pi(t) (i = 1, . . . ,n) (aij = const). dt As in Section 9.5, we replace this system by the vector equation - = ailxl

where A is now a constant n x n matrix and p(t) is a continuous vector function. For simplicity we take a = -00 and b = 00 and consider all solutions in the interval -00 < t < 00. As in Section 9.5, the general solution of the homogeneous equation

is given by X

+

= C I X I ( ~ ). . . f ~ n ~ n ( f ) ,

provided that xl(t), . . . , xn(t)are linearly independent solutions. To find such solutions, we seek solutions of form x = e A t u,

(9.40)

where h is a constant scalar and u is a constant nonzero vector. Substitution in (9.39) gives heAtu= A

~ ' ~ U or

Au = Au.

Thus (9.40) is a solution precisely when u is an eigenvector of A, associated with the eigenvalue A. By the theory of Section 1.1 1, A must be a root of the characteristic equation: det (A

- h l ) = 0,

(9.41)

Chapter 9 Ordinary Differential Equations and each real root A of this equation is an eigenvalue. If there are n distinct real eigenvalues Al, . . . , A,, then we obtain n corresponding eigenvectors u l , . . . , u, and n solutions A r (9.42) X I (t) = eA"ul , . . . , x,(t) = e u,. Furthermore, the eigenvectors u l , . . . ,un are linearly independent vectors (Problem 12 following Section 1.16), and hence the solutions (9.42) are linearly independent solutions of (9.39) (Problem 6 which follows). Therefore the general solution of (9.39) is (9.43) x = cleA1'ul + . . . + c,eLrun. EXAMPLE1

[-i -:I,

$ = ~ - 2 y , ~ = - 2 x + ~ , o r ~ = [ - i- ~ ] x . ~ e r e x ~ = x , x 2 = y .

.:1'

A= the characteristic equation is ,l2*1 = 0 or i2- 2A - 3 = 0, and hl = 3, A2 = - 1 are eigenvalues with associated eigenvectors ul = (1, - l), u2 = (1, 1). The general solution is )ti

or in the original notations, ~ = c ~ e ~ ' + c ~ e -y~= -, c l e -

31

+ c2e-'.

If the eigenvalues i l , . . . , A, are distinct but some are complex, then one obtains n linearly independent solutionsby taking real and imaginary parts of the corresponding expressions eA'ufor complex roots. (See Section.6.3 of ODE for a proof.) EXAMPLE 2

2=

x. Here we find the eigenvalues to be 1, fi.Corre-

sponding to 1 we obtain the eigenvector u l = (1, 1, 1) and the solution xl(t) = erul. Corresponding to +iwe obtain the eigenvector u = (1, i, - 1) and the solution .,."

+

e i , [ ~ ] = [ ~ r ] = [ c O s-sin t ' i st i ni ]cos ~ [tc O s t ]-sin t

-1

-cos t - i sin t

-err

Hence we obtain the real solutions x ~ ( t= )

[

-COS

[ [

+i

t

cost . -sin t .,,

sint] x3(t) = cos t . -sin t

-sin C O S t~ ] , -cos t

[:I

f

8.

[

(The complex root -i leads to the same two solutions.) The general real solution is x = ~ 1 x(t) 1

+ c2x2(t)+ c3x3(t)=

~1

e'

cost]

+ c2 -sin r + c3 -cos t

This can be verified directly (Problem 4 which follows). ...-

sint]

cost -sin t

.

641

642

Advanced Calculus, Fifth Edition When repeated roots occur, the procedure has to be modified. If A is an eigenvalue of multiplicity k, one replaces the expression eh'u by eAtq(t)where q(t) is a column vector function whose components are polynomials of degree at most k - 1. It can be shown that n linearly independent solutions can be found in this form (with complex roots treated as previously). For details, see Section 6.22 of ODE. The method of Laplace transforms also handles all cases without difficulty (see Problem 3 which follows). Nonhomogeneous equations. As in Section 9.5, we obtain a particular solution x*(t) by the variation-of-parameters formula (9.37). The general solution is then obtained as in Section 9.5 as 2

9

2

+

,

:

+

EXAMPLE 3 = x - 2 y cost, = -2x y - sin t. The related homogeneous system is that of Example 1, and therefore we can take

We also find

X-'(t)p(t) d t to be

'/ ['-"I';: 2

et

I,:[

1

dt = 2

[11

+

e-3' (cos t sin t) d t er(cost - sin t) dr

1

and therefore find

-

The general solution is

x = cle3'

I + c2e-' + -(3 10

y = -cle

3t

1 + c2e7' + -(7 10

cos t - sin t ) , cos t

+ sin t).

7

Chapter 9 Ordinary Differential Equations

PROBLEMS 1. Find the general solutions:

Find the general solution of the system:

by solving the equivalent system: dx dy dz - =x-4y, - = W, -= z , dt dt dt b) Find the general solution of the equation

.&

dw -=y-x. dt

by solving the equivalent systeni: --.

3. One may be able to obtain the solution of (9.38) for t 2 0 with initial value xo for t = 0 by applying Laplace transforms as in Problem 16 following Section 9.4. For Example 1 in the text, let (x(t), y(t)) be the solution with initial value (1,2) and let 49[x(t)] = X(s), Y[y(t)] = Y(s). Then we obtain sX(s) - 1 = X(s) - 2Y(s), sY(s) - 2 = -2X(s) Y(s). We solve these simultaneous equations for X(s), Y(s), obtaining X(s) = (3)(s I)-' (f)(s - 3 ) ~ ' . ~ ( s ) = ( f ) ( s - 3)-' + ( i ) ( s I)-', so that x =($e-' -(f)e3'. y =(&)e3' (%)e-'. Apply the method to the following equations with initial conditions. [The rule 2'[tkea'] = k!(s - a)-k-' will be found helpful. In parts (e) and (f) there are multiple eigenvalues.] a) Equations of Problem l(a), with x(0) = 2, y(0) = -1. b) Equation of Problem I(b), with x[(O) = 3, x2(0) = 2. C) Equation of Problem l(c), with x(0) = 0. d) Equation of Problem l(d), with x(0) = col(1, 1, 1). . -

+

+

+

+

4. In the text a general solution for Example 2 is obtained. Verify that this is indeed the general solution by showing that the vector functions xl(t), x2(t), x3(t) are linearly independent for -00 c t < oo and that each function is a solution of the differential equation for -00 < t < 00. 5. For a matrix function A(t) = (aij(?)),a 5 t 5 b, one defines the derivativedA/dt to be the matrix function (daij/dt), provided that all aij(t) have derivatives. One defines 1 A(?)d t to be the matrix ( jaij(t) d t ) and S,b A(?)d t to be (hb aij(t) dt); here we assume all aij(t)

Advanced Calculus, Fifth Edition to be continuous over the interval considered. Verify the following rules for appropriate hypotheses on the sizes of the matrices X ( t ) , Y ( t ) ,and on the continuity or differentiability of their elements.

+

a) ad ( ~ ( t ) Y O ) ) =

% + g;

+

b) $ ~ ( t ) ~ = ( t ~) ( t ) g~ 0 1 % ; C)

If X ( U ) = (x,,(u)) and u = + ( t ) ,then

+

hb

+

9= 4 ' ( t )d=X; '9

Y(t)dt. d) h b [ x ( t ) Y ( t ) ] d t= X ( t ) d t 6. Prove: If ul , . . . , u, are linearly independent vectors of V n ,then the solutions (9.42)of (9.39)are linearly independent for -m < t < oo. 7 . a ) Show that the solutions (9.34)can be written as x = X(t)c,where c = col (cl, . . . ,c,). b) Show that if X ( t ) is chosen as in the paragraph following Eq. (9.35),then the solutions (9.34)can be written as x = X(t)x(to). 8. Prove: If cl, . . . , c, are constants, xl(t), . . . , x,(t) are linearly independent solutions of Eq. (9.31 ) for a < t < b, and x*(t)is a solution of the nonhomogeneous equation (9.30') for a < r < b, then Eq. (9.35)defines x = x(t) such that x(t) is a solution of the nonhomogeneous equation (9.30') for a < t < b, and every solution of the nonhomogeneous equation for a < t < b can be represented in the form (9.35).Thus (9.35)gives the general solution of Eq. (9.30').

In many physical problems, one is led to simultaneous differential equations of the form

Here ml, . . . , m, are positive constants (typically, masses), V(xl, xz, .. . , x,) is the potential energy of the system, and V has a minimum at (0, . . . ,0). Hence a Vlax, = 0 at (0, . . . , 0 ) for i = 1, . . . , n, and if we expand V by Taylor's formula (Section 6.21), we obtain an expression of form

where A = (a;,) is a symmetric matrix. We assume further that the quadratic form C ai,x;xj is positive de$nite, which ensures that V has a minimum at (0, . . . ,0); accordingly, all eigenvalues of A are positive (see Section 2.2 1).To study the behavior system of the system near (0, . . . , 0), we drop the higher-degree terms and replace V by

in the differential equation. Since aV/axi = C a i j x j , the new equation can be written in vector form as follows:

Chapter 9 Ordinary Differential Equations Here we make a change of variable, setting y = Nx, so that

where B = N-I AN-'. Since A is symmetric, we verify that B is also symmetric (Problem2 below). Therefore B is similar to a diagonal matrix D = diag (A1,. . . , A,), where A l , . . . , An are the eigenvalues of B, which are also positive (Problem 2 below); we write -

B=c-'DC,

19

sothatD=C~C-'.

If we now let y = C-'z, then we obtain z = Cy and

that is

Since each A, is positive, we can write Ai = o;, and the solutions of (9.47) are given by zi=A,sin(wit+a;), i = l , ..., n , (9.48) where A,, . . . , A,, al, . . . , a, are arbitrary real constants. Now x = N-' y = N-'C-'z = Ez, for an appropriate matrix E = (el,). Hence n

n

This is the general solution of (9.46). If, for example, A2 = . - . = A, = 0, then x; = eilA1sin(olt + a l ) , i = 1 , . . . , n ,

(9.50)

and all coordinates x, oscillate synchronously with frequency o l . This is called a nornzal mode of oscillation of the system. The general solution (9.49) can be regarded I as a superposition of normal modes. ' "" For physical examples of such vibrations, see Sections 10.3 and 10.4.

[!:

;:I,

Here A = and A has eigenvalues 16 and 4, so that V is positive definite and has a minimum at (0,O). To find the solutions, we can simply set xi = u, sin (At + a , ) in the differential equations, since we know that there are solutions of this form-the normal modes. We obtain the equations

+

(10 - 4h2)ul - 6u2 = 0,

- 6 ~ 1 (10 - 4h2)u2= 0.

.)

They are satisfied for h = f1, h = f2. For h = 1 we obtain ul = 1, u2 = 1; for A = 2 we obtain ul = 1, uz = 1. Hence we obtain the normal modes

-

646

Advanced Calculus, Fifth Edition

+

+

x = sin ( t arl)(l, 1) and x = sin (2t aZ)(lr-1). The general solution is an arbitrary linear combination of these normal modes:

+ a1) + c2 sin (2t + a2), x2 = CI sin ( t + a l ) - cz sin (2t + ad. xl = cl sin (r

(We remark that the choices h = - 1 and h = -2 yield no additional solutions.)

PROBLEMS 1. Find the normal modes of oscillation:

2. Show that the matrix B = N - ~ A N - ' of Section 9.7 is symmetric and that the associ~ted quadratic form b i j y i y j is obtained from C aijxix, by replacing each xi by y i / m , ! , so that b;,y; yj is also positive definite.

The Taylor series f(a)+

f '(a)(x - a )

+

f "(aXx - a)?

f '")(a)(x - a)" + ... ..-- ' n!

2! (Section 6.16) of a function f (x) can be formed when all its derivatives at x = a are known. If, for example, f satisfies the differential equation + . . a +

y1 = F(x, y) with given initial condition f (a) = yo, then the equation itself gives y' = f '(a). Thus f'(a) = F[a, f (a)]. If F has continuous derivatives for the values of the variables concerned, then one can differentiate to obtain a formula for the second derivative: ,, aF aF y = - -y', ax ay

+

so that

By proceeding in this way, one obtains a definite Taylor series for f .The fact that the series obtained does converge in an interval about x = a and represents a solution y = f (x) of the differential equation can be justified under suitable assumptions about the function F(x, y). In most applications, F is a rational function of x and y, and the method is applicable except when the denominator is zero at (a, yo). More generally, the method is applicable when F is analytic, that is, can be expanded in a power series in powers of x -a and y - yo in a neighborhood of (a, yo) (Section 6.20). For proofs the reader is referred to Chapter 12 of ODE.

647

Chapter 9 Ordinary Differential Equations EXAMPLE

y' = x 2 - y 2 ; y = 1 for x = 0. Here

-.

Hence at x = 0, one has, for the solution sought,

Thus the solution is given by

It is essentially hopeless to attempt to find the general term here; this is a characteristic defect of the method when it is applied to a nonlinear differential equation. However, the general theorem referred to earlier gives assurance that the series does converge for x in some interval -a < x < a and is a solution. Furthermore, it is possible to obtain estimates for the interval of convergence and for the remainder after n terms. Hence the series can be used for carefully controlled numerical work. The method described applies equally well to higher-order differential equations and to systems of equations. Thus for the equation

with initial conditions y = 1 and y' = 2 for x = 1, one has y"' = 2yy'

+ I,

y" = 2y'2

+ 2 y y " , ... ,

so that

For linear differential equations with variable coefficients, another way of obtaining the series is available and will often make it easy to find an expression for the general term of the series. This is a method of "undetermined coefficients." For example, let the equation be y"+xyf+y

=o,

and let the solution be sought in the form of a series about x = 0:

Upon substitutingthis series in the equation and collecting terms according to powers of x , one obtains the equation:

By the corollary to Theorem 40 of Section 6.16, each coefficient must equal 0. Hence one obtains the equations:

,

648

Advanced Calculus, Fifth Edition Hence there is a recursion formula for the coefficients:

Thus c 2 = - - co , 2

Cq=

co 2.4

c6=--

co 2.4.6'""

It appears that co and c, are arbitrary constants; they are, in fact, simply the initial values of y and y' at x = 0. The solutions found can now be written in the form:

Here an application of the ratio test shows that the series converge for all x . Furthermore, y satisfies the differential equation for all x. Since the functions

-

r

are clearly linearly independent, the functions

actually give the general solution of the differential equation. This method can also be used for nonlinear equations, but the algebraic processes usually become highly involved, and the chances of obtaining the general term of the series or an expression for the general solution are very small. For many applications it is important to have a series solution of a differential equation y' = F ( x , y ) , even when the function F ( x , y ) is not analytic in a neighborhood of the initial point considered. The most common case is that in which F is a rational function whose denominator is 0 at the initial point. The initial point is then a singular point of the differential equation, and no general statement can be made about solutions. However, it is possible in many cases to obtain the solutions through and near the singular point in the form of series of appropriate types. For example, the series where m is not necessarily positive or an integer, can be used in certain cases. In other cases the solution can be expressed as a series

-

,

where m and p are quite general. Similar remarks apply to equations of higher order and to systems of equations. Series of type (9.51) are of special importance for linear equations. For example,

649

Chapter 9 Ordinary Differential Equations Bessel S differential equation of order n:

it

I

.

has solutions of this form (with m = n). One solution is the function Jn(x), the Bessel function of order n, considered in Section 7.15. Further examples are given in Problems 7 to 9 which follow. For nonlinear equations, one can often express the solutions near the singular point by means of series of type (9.52) in terms of a parameter t. For a full discussion of this subject, see the books of Ince, Picard, and Whittaker and Watson listed at the end of this chapter. It will be found that the theory of functions of a complex variable is essential for a complete analysis of the problem. Numerical methods. Approximate methods for numerical evaluation of solutions have long been known. With the development of the digital computer they have been greatly improved and refined. For further information, see the books of Henrici, Hildebrand, Iserles, and Forsythe, Malcolm, and Moler listed at the end of the chapter. The numerical methods have a close connection to the series solutions of this section. Each method can in fact be interpreted as obtaining solution values on the basis of a certain number of terms in a series solution.

PROBLEMS 1. Obtain the first four nonzero terms of the series solution for the following equations with

initial conditions: a) y f = x 2 y2 + 1 , y = l f o r x = l ; b) y' = sin(xy) + x2, y = 3 for x = 0; c) y " = ~ * - ~y~= l,a n d y f = O f o r x = O ; d) yfr'=xy+yyf, y = O , y l = 1 , a n d y M = 2 f o r x=O. 2. Obtain the general solution in series form: a) y" + 2xy' + 4y = 0 b) y" - x~~ = 0 3. Obtain a solution such that y = 1 and y' = 0 for x = 0 for the equation: yfr+ y' +xy = 0. 4. Obtain in series form, up to terms in x3, a solution of the system

such that y = 1 and z = 0 for x = 0. 5. Show that

satisfies Bessel's equation of order 0 in the form

for all x.

.i

650

Advanced Calculus, Fifth Edition

6. Determine, if possible a solution of Bessel's equation of order 1: ~ 2 ~ ' 'xy' (x2 - l)y = 0, h&ng the form

+ +

7. Determine, if possible, a solution of Bessel's equation of order n having the form

8. a) Determine a solution of the hypergeometric equation:

+ + +

+

(x2 - x)yl' [((Y /3 1)x - yly' aBy = 0 (a,#yI,constants), having the form given in Problem 7. b) Determine a solution of the hypergeometric equation having the form

9. a) Find solutions of LRgendreS equation: (1 - x ~ ) ~-' 2xy' '

+ + l)y = 0 having the U((Y

form

b) Show that the Legendre polynomial Pn(x)(Section 7.14):

is a solution when Pn(x). [Hint: Set

(Y

= n and that every polynomial solution is a constant times a

z = (x2 - 1)"; show that (x2 - 1 ) ~= ' 2nxz. Differentiate both sides n 1 times with the aid of the Leibnitz Rule of Section 7.14.1 c) By regarding the Legendre polynomial as a particular power series solution, obtain the formulas (for n 2 1):

+

+.. . +(-l)l , , ( n + l ) ( n + 3 ) - . . (n!2 n -

l)n(n-2).-.2

.'I,

n even;

651

Chapter 9 Ordinary Differential Equations

9.9 THEEXISTENCE AND UNIQUENESS THEOREM

'

We consider the general system of first order equations

and the linear system

THEOREM 5 I. Let D be an open region in the (n + 1)-dimensional space of coordinates (t, X I ,. . . ,x,). Let the functions Fl in (9.54) be defined and continuous in D, with continuous first partial derivatives aFi/axj. Then for each point (5, El, . . . ,3,) in D there exists a solution

of the system (9.54) for which &(i) =R;

( i = 1, ..., n).

(9.57)

11. In the system (9.55), let the functions aij(t) and fi(t) (i, j = 1, . . . , n) be defined and continuous for a 5 t 5 b, let a 5 i 5 b, and let Z1,. . . , E n be given numbers. Then there exists a solution of the system (9.55) for which (9.57) holds. 111. The solution found in I and I1 is unique; there is no other solution of the same system in the interval stated satisfying the same initial conditions (9.57). Proof. I. We choose an (n

+ 1)-dimensional rectangle

in D. Let (t, xi, . . . , x:), (t, x;', theorem (Section 6.2 1),

. . . ,x:)

be two points in R l . By the Mean Value :

where (t ,x;, . . . ,x,') lies on the line segmentjoining (t, xi, . . . ,x:) to (t, x;', . . . ,xi) in R1. Since the first partial derivatives are continuous in R 1 , we can choose K so large that t <

K

in R1 (i, j = 1, .. .,n).

Hence from (9.59), for i = 1, . . . ,n,

(9.60)

Advanced Calculus, Fifth Edition An inequality such as (9.61) holds for each cldsed rectangular region in D; one says that the F; satisfies local Lipschitz conditions in D. It is these conditions that are used in the proof rather than the hypothesis that the aFi/axj are continuous. Since the Fj are continuous, we can also choose M so that for i = 1 , . . . , n,

We now choose h so that O
k

h l Ml . . . . ' h 5

Henceforth we restrict attention to the (n

-.kn

'

(9.63)

M

P + 1)-dimensional rectangle (contained B$'#lh"

Our solution is now defined as the limit of a sequence of successive approximations (method of Picard):

4,(0)(t) = 2;

(i = 1, . . . , n),

and inductively,

for N = 1,2, . . . . Throughout, It - il 5 h. We must verify that the 4fN'(t) are well defined. This we do by showing inductively that at each stage the graph of (N) xi = 9; (t) (i = 1, . . . , n) in the (n 1)-dimensional space lies in R. This is clearly true for N = 0.If it is true for N = k, then Eq. (9.65) is meaningful for N = k 1 and defines the functions +!k+"(t). Furthermore,

+

+

+

by (9.62) and (9.63). Hence the graph lies in R for N = k 1, and by induction the 4iN'(t)are well defined for all N. By their definition the +!N'(t) are continuous for It - T I 5 h. We now show that they converge uniformly to the desired functions &(t) (i = 1, . . . , n). We have

=

16'

F; (u, @)(u), . . . ,4F)(u))du

Chapter 9 Ordinary Differential Equations

by (9.61). Proceeding inductively, we find that

The series of constants

MN converges (by the ratio test), and hence the series

converges uniformly for It - 71 a h to a continuous function 4i(r) (Section 6.13). The Nth partial sum of the series is

Hence the sequence 4,'N'(t) converges uniformly to 4,(t). Since the graph of xi = 4,'N'(t)(i = 1, . . . ,n) is in R for each N, we conclude that the graph of xi = #,(t) (i = 1, ..., n)isin R. Furthermore, by (9.61), n

1 6(t, 4iN)(t),. . . 4AN)(t)) - F,(t, 4l(t), - 4n(t))/ 5 K 9

9

,=I

14y)(t)- 4J(t)l.

Accordingly, for each i the sequence F,(t, +iN'(t), . . . , +AN'(t))converges uniformly to F,(t, +l(t), . . . , @,(t)). Hence we can let N + oo in Eq. (9.65) to obtain

Accordingly, for It

- 71 < h,

and #,(i)= 2,. Thus x, = &(t) (i = 1, . . . , n) is the solution sought. 11. (Linear case). Here aF,/ax, = a,,(?). By continuity we can choose K so that Ja,,(t)l 5 K for a _( t 5 b. Thus (9.61) holds in the infinite closed region R : a ( t ~ b , -oo<x;
for i = l ,

...,n.

a%

.ci,

653

654

Advanced Calculus, Fifth Edition We can proceed as before, taking $,'O'(t)= 2, for a 5 t 5 b and defining $ y ' ( t ) by (9.65).These functions are then well defined and continuous for a 5 t 5 b, with graphs in R. We let M be chosen so that

and have again

We proceed as before to obtain (9.66),choosing h so that b - i 5 h , i - a 5 h . The rest of the argument is the same, and we obtain the desired solution & ( t ) . It should b, and hence (9.68)follows, with be remarked that now (9.67)is valid for a 5 t I the derivatives at t = a , b understood as derivatives to the right and left, respectively. 111. (Uniqueness). We return to the general case of I. .Let us suppose that we have found a second solution x i = + i ( t ) (i = 1 , . . . , n), It - il < h with + ; ( i ) = X i . Since this is a solution of (9.54), ( + ( ( t ) ( = I Fi(t, + l ( t ) ,. . .)I 5 M as long as the solution remains in the rectangular region R. Since k; 5 M h by (9.63),it follows that the solution cannot leave R for ( t - il < h (Problem 8 following Section 9.9) and must therefore have a graph in R. From (9.54)we can write

and hence from (9.65),

Also

since IF, ( t , X I , . . . , x,)l 5 M in R. By induction as before, using (9.61), we conclude that for It - il < h ,

%'

The series MN converges, so that M N +- 0 as N +- m. Therefore the sequence $IN'(t)converges uniformly to +,( t ) for It - il < h . But we already know that ~ $ , ' ~ 'converges (t) to & ( t ) on this interval. Therefore $ , ( t ) = + , ( t ) for Jt - t J < h, as asserted.

Chapter 9 Ordinary Differential Equations In the linear case we proceed similarly, with R as in Case 11, so that x; = # ; ( t ) and xi = + i ( t ) (i = 1 , . . . , n , a 5 t 5 b ) have graphs in R. We let M be chosen so that I Fi(u,lCrl(u),. . . , +,(u))l 5 M for a 5 u 5 b , i = 1 , . . . , n. Then as before, we conclude that ~ $ , ' ~ ' converges (t) to q i ( t ) , so that C$;(t)= lCri(t)for a It < b.

a Remarks. For the general system (9.54)we have now proved "local" existence and uniqueness-that is, we have shown that there is a unique solution through the initial point in a small interval It - il < h. If the solution approaches as limit a point of D as t + i h (or as t + i - h ) , then there is a unique solution through that point, and we verify that it provides an extension of the previous solution to a larger interval, say, i - h < t < i h 6 (6 > 0). If this solution has a limit as t + i h + 6, we can repeat the process. In this way we can prolong the solution for increasing and decreasing t and eventually obtain a solution in an interval a! < t < j3, where a! may be -00 and j3 may be +m, which cannot be prolonged. We call this a complete solution of the system (9.54).There is thus a unique complete solution through each point (7, Rl, . . . , i nof) D.

+

+ +

+

One can show further that each complete solution must approach the boundary of D as t -+ a or t + p ; here we include receding to infinite distance ( t 2+ x : . . + x i + o o ) as a form of approach to the boundary of D (see Section 12-4 of ODE). One can also consider a linear system (9.55) in a domain D: a! < t < j3, -oo < x; < m for i = 1, . . . , n , so that the t-interval is open. For each closed interval a I: r 5 b contained in this interval we obtain a solution through (i,f l , . . . , 1 , ) as before, ending at a point of D. It follows that the corresponding complete solution is defined for a! < t < /?.For a nonlinear system in such a domain D this may fail, as is illustrated by

+

dx -=I+x2, -00
where p, q and f are continuous for a Ix 5 b . We replace x by t and consider the corresponding system

655

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Advanced Calculus, Fifth Edition

.

)

By Theorem 5 there is a unique solution xi = & ( t ) (i = 1,2) of the system (9.70)with, say, xl = 0 , x2 = 0 for t = i, where a 5 i 5 b. Thus

Hence

If we replace r by x , 4 1 ( t by ) y*(x),& ( t ) by y*'(x),then

Thus y = y*(x) is a particular solution of the nonhomogeneous equation (9.69). In the same way we prove the existence of solutions of the related homogeneous equation. If we choose two solutions of (9.70) such that the corresponding $matrix X ( t ) of Section 9.5 is nonsingular, then as in Section 9.5 the solutions are linearly independent; we verify that this implies that the corresponding functions yl ( x ) , y2(x) are linearly independent solutions of the related homogeneous equation. For, from the form of (9.70),

where det X ( t ) # 0 . Now, as before, yl(x) = $ , ( x ) ,y2(x)= 4 2 ( x ) are solutions of the related homogeneous equation for (9.67). If, with cl and c2 not both 0,

+

clyl(x) c2y2(x) = 0 ,

a Ix I

b 7

then also, by differentiation,

Hence the Wronsluan determinant

must be identically 0. But if we return to t , X I , x2,the determinant is simply det X ( t ) , which we know is never 0 . Thus y l ( x ) ,y2(x)are linearly independent, as asserted.

PROBLEMS 1. For each of the following differential equations with initial conditions, show that the solution exists in the interval given and apply the method of successive approximations to @ ( 2 ) ( t )4, ( 3 ) ( ~ ) . obtain @ ( O ) ( t ) r n, $(')(t), d x 2 a) - x + t 2 , 5 = 0, 2 = 0, It1 < 0.5 [Hint: The maximum of (FI in a rectangle It 1 5 h, Ix 1 5 k is M = h2 k2. As in the proof of Theorem 5, we must choose h so that h 5 k / M = k / ( h 2 k 2 ) . Show that for appropriate k this permits h to be 0.5.1

+

+

Chapter 9 Ordinary Differential Equations

2. a) Show that F(t, x) = tlxJ is continuous for all (t, x), that F, exists and is continuous except for x = 0, and that F satisfies local Lipschitz conditions for all (t, x) (as in i (9.61) with n = 1). b) Find all solutions of the differential equation d x l d t = tlxl and show that a unique solution passes through each point. 3. a) Show that F(t, x) = x4/' is continuous and has a continuous derivative F, for all (t, x). b) Find all solutions of the differential equation d x l d t = x4I3 and show that there is a unique solution through each point. i 4. a) Show that F(t, x ) = x1I3 is continuous for all (r, x) but that F, does not exist for x = 0 and that F(t, x) fails to satisfy a local Lipschitz condition for all (t, x). b) Obtain all solutions of the differential equation dx/dt = x113and verify that there is more than one solution through each point on the t-axis. 5. For each of the following differential equations, find x = +(t; 7, x), the complete solution through (7, f ) , and determine whether x = +(t; 0, f ), gives all solutions:

a)

- 2x

C)

= e'x2,

e)

dr

+

= e',

2 = t sinx, d) 2 + ,*yx2 =0

b)

(x > 0).

= 0, all (t. x) except (0,O).

6. Let the following system, with initial condition, be given: dx i -=tx-y, 9=tx2-y, x = l and y = l f o r t = O . dt dt a) Show that the solution can be obtained for It 1 < 0.3 by successive approximations. b) Ohtain the first three successive approximations for x and y. c) Obtain the power series solution through terms in t3 and compare with the result of part (b). 7. Obtain the complete solution satisfying the given initial conditions: a ) y" = evy', y = 0 and y' = 1 for x = 0; b) d x / d t = y c o s t + t + x , d y / d t = 1 + y 2 , x = O , y = O f o r t = O ; c) d x l d t = 4x - 5y, dy/dt = x - 2 y , x = xo, y = yo fort = to. 8. a ) Let x = $(t), It - 71 p h , $(i) = 2, where $'(t) is continuous and it is known that I$'(t)( 5 M whenever (t, $(r)) is in the rectangle R: It - il p h , Ix - f 1 5 k, where Mh 5 k. Show that (1, $(t)) is in R for It -TI 5 h.

+,

b) Generalize the result of part (a): Let x, = (t), It - t 1 ( h , $,ti) = f , ( i = 1 , . . . , n), where +,'(t) is continuous and it is known that I$:(t)l ( M whenever (t, $ ~ ( t ) ,. . . , &(t))isintheregionR: It-TI 5 h , Ix, -2,I 5 k, (i = 1 , . . . , n),where Mh 5 k, for i = 1, . . . , n:Show that (t, $l(t), . . . , $,(t)) is in R for It - il 5 h . 9. Show that each complete solution of the differential equation dx - = t2arc tan x + e' dt is defined for -co < t < co.(See Problem 8(a).)

Suggested References Agnew, Ralph P., Differential Equations, 2nd ed. New York: McGraw-Hill, 1960. Andronov, A. A., S. E. Khaikin, and A. A. Vitt, Theory of Oscillators. Reading, Mass.: Addison-Wesley, 1966.

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658

Advanoed Calculus, Fifth Edition Coddington, Earl A., and Norman Levinson, Theory of Ordinary D8erential Equations. New ' York: McGraw-Hill, 1955. Forsyth, A. R. Theory of Differential Equations, 6 vols. New York: Dover, 1964. Forsythe, George E., Michael A. Malcolm, and Cleve B. Moler, Computer Methods for Mathematical Computations. Englewood Cliffs, N . J.: Prentice-Hall, 1977. Goursat, ~ d o u a r dCourse ,~ in Mathematical Analysis, Vol. 11, Part I1 (transl. by E. R. Hedrick and 0. Dunkel). New York: Dover, 1964. Henrici, Peter K., Discrete Variable Methods in Ordinary Differential Equations. New York: John Wiley & Sons, 1962. Hildebrand, F. B., Introduction to Numerical Analysis, 2nd ed. New York: McGraw-Hill, 1974. Ince, E. L., Ordinary Differential Equations. New York: Dover, 1956. Iserles, Arieh, A First Course in the Numerical Analysis of Differential Equations. Cambridge: Cambridge University Press, 1996. Kamke, E., Differentialgleichungen reeller Funktionen. Leipzig: Akademische Verlagsgesellschaft, 1950. Kamke, E., Differentialgleichungen,Losungsmethoden und Liisungen. Leipzig: Akademische Verlagsgesellschaft, 1961. Kaplan, Wilfred, Ordinary Differential Equations. Reading, Mass.: Addison-Wesley, 1958. McLachlan, N. W., Ordinary Non-Linear Differential Equations in Engineering and Physical Sciences, 2nd ed. Oxford: Oxford University Press, 1956. Picard, Emile, Trait6 d'Analyse (3 vols.). Paris: Gauthier-Villars, 1922. Rainville, Earl D., Intermediate Differential Equations. New York: John Wiley & Sons, 1943. Ralston, Anthony, and P. Rabinowitz, First Course in Numerical Analvsis. 2nd ed. Mineola, New York: Dover, 200 1. Scarborough, James B., Numerical MathernaticalAnalysis, 6th ed. Baltimore: Johns Hopkins Press, 1966. Whittaker, E. T., and G. N. Watson, A Course of Modern Analysis, 4th ed. Cambridge: Cambridge University Press, 1950.

Partial Differential Equations

A partial differential equation is an equation expressing a relationship between an unknown function of several variables and its derivatives with respect to these variables. For example,

are partial differential equations. By a solution of a partial differential equation is meant a particular function that satisfies the equation identically in a domain of the independent variables. For example,

u = eTt sinx,

u = x 2 - Y2

are solutions of (10.1) and (10.2), respectively. Although partial differential equations were not studied as such in the earlier chapters, they occurred in a number of important connections. For example, the solutions of (10.2) are precisely the harmonic functions of x and y; such functions were studied in Chapter 2. See also Sections 5.16 and 8.11. The Cauchy-Riemann equations u, = v,, u, = -v, form a system of partial differential equations; they

Advanced Calculus, Fifth Edition were studied in Chapter 8. Other systems encountered earlier in this book are the following:

The first and second sets arose in consideration of line integrals in the plane and in space (Sections 5.6 and 5.13); the third set occurred in the study of solenoidal vector fields (Section 5.13). The results of Section 2.22 also concern partial differential equations, formed from Jacobians. In this chapter we shall not attempt to consider general methods of determining the solutions of partial differential equations, but shall rather confine ourselves mainly to one class of linear partial differential equations-namely, equations of the form

where u is a function of t and one, two, or three space variables x , . . . ; p , H, and K 2 depend on the space variables, and F depends on t and the space variables. This equation is a natural generalization to continuous media of the equation

for forced vibrations of a spring. It will be seen that the parallel between (10.3) and (10.4) is far-reaching. In order to make clear the relationship between (10.3) and (10.4) we shall first study the case of the motion of nyo particles attached by springs. The results obtained will then be generalized to the case of N particles. It will be seen that the equations governing the motion form a system of form

the quantities u I , . . . , u,, are coordinates measuring the displacements of the various particles from their equilibrium positions. The expression in brackets [. . .] depends on the u, and, in particular, on their diferences: u2 - u I, u3 - uz, . . . . We shall show that such general systems are capable of "harmonic motion," damped vibrations, exponential decay, and forced motion, just as is the single mass governed by (10.4). If we let N become infinite, then n becomes infinite in (10.5), and we obtain as a "limiting case" precisely the partial differential equation (10.3). The differences in the expression [. . .] turn into the partial derivatives out of which V2u is built. Finally, it will be seen that the partial differential equation obtained as limit continues to exhibit all the properties observed for the systems of 1 , 2 , . . . , N particlesharmonic motion, exponential decay, and so on. The one basic difference between the various cases is that whereas for one' particle there is only orze frequency of oscillation, for two particles moving on a line there are two frequencies, for N particles moving on a line there are N frequencies, and for infinitely many particles there are infinitely many frequencies.

Chapter 10 Partial Differential Equations The discussion to follow applies the theory of linear differential equations as developed in Sections 9.4 to 9.7. FOR FORCED VIBRATIONS OF A SPRING 10.2 REVIEWOF EQUATION

We recall briefly some basic facts concerning Eq. (10.4). Throughout we assume m = O , h = O , k >0.

a ) Simple harmonic motion. Here h = 0 , F ( t ) = 0 , and m > 0. The equation becomes

The solutions are sinusoidal oscillations:

b) Damped vibrations. Here m > 0, F ( t ) = 0 , and h > 0, but h,is small: h2 < 4mk2.The equation becomes

The solutions are oscillations with decreasing amplitude:

where a = h /2m and ,!3 = (4mk2 - h2):/2m. c) Exponential decay. Here m = 0,h > 0,F ( t ) E 0 . The equation reads

The solutions are decaying exponential functions: >

k' ~ = c e - ~ ' a, = h'

(10.11)

A similar result is obtained if we consider an equation (10.8) in which h is large in comparison to m: h2 > 4mk2; in fact, (10.10)can be considered as the limiting case: m -+ 0 of (10.8). d) Equilibrium. We assume F ( t ) = Fo, a constant. The equation becomes

The equilibrium vaiue of x is that for which x remains constant; hence d x l d t = 0 , d 2 x / d t 2 = 0.Accordingly, at equilibrium,

661

662

Advanced Calculus, Fifth Edition

"'

'

e ) Approach to equilibrium. To study the approach to equilibrium, we let so that u measures the difference between x and the equilibrium value. We find that

s o that u has the form (10.7), (10.9), or (10.1 1). For example, when m = 0 and h > 0 [case (c)],

x approaches its equilibrium value exponentially.' f ) Forced motion. We allow F to be a general function of t. If, for example, m = 0, h > 0, then

Each solution consists of a transient plus a particular solution:

x * ( t ) = -e-ar

h

S

earF ( r ) d t .

(10.20)

The solutions attempt to follow the "input" ~ ( t ) / k are ~ , hindered by the friction. Remark. The equations for which pn = 0 can be realized by other physical models-for example, the cooling of a hot mass or an electric circuit containing resistance and capacitance.

We consider the model illustrated in Fig. 10.1. Two particles of masses m I , m 2 are attached to each other and to "walls" by springs. The particles move on an x-axis and have coordinates xl, XZ; the walls are at xo,xs. For simplicity we assume that all springs have the same natural length 1 and the same spring constant k2. We assume

Figure 10.1

Linear system of two masses.

Chapter 10 Partial Differential Equations the particles to be subject to resistances -hl(dxl/dt), -h2(dx2/dt) and to outside forces Fl(t), F2(t). The differential equations have fhr: form

They simplify to the following:

We have not explicitly stated that the walls are fixed at xo, x3, and the differential equations remain correct even if the walls are moved in a manner controlled from outside the system. However, if .xu and x 3 are constants x,*,x; and Fl(t), F2(t) are 0, the system has precisely one equilibrium state, namely, the solution of the equations

The solution is at once found to be

at equilibrium the particles are equally spaced between the walls. We now refer each particle to its equilibrium position by introducing new variables:

as suggested in Fig. 10.2. The differential equations (10.21) then have the appearance

When the walls are fixed, xo = xg*,x3 = x ; , uo and u3 are 0; however, (10.25) allows for a general motion of the walls and, in particular, for constant nonzero values of uo and 142, which would signify a shift of the equilibrium position of the walls. We now consider several special cases of (10.25), paralleling the cases (a), (c), (d), (e), and (f) of the preceding section. The discussion of the analogue of (b) is left to Problem 9. of f

Figure 10.2

Displacements from equilibruun positions,

,

.

, , ,r

* A

481

i

663

664

Advanced Calculus, Fifth Edition

a) Harmonic motion. We assume m l > 0, m2 > 0, hl = h2 = 0, uo = uj = 0, and F l ( t ) = F2(t)= 0, SO that the differential equations (10.25) become

These are equivalent to four first-order equations:

and can therefore be solved by exponential substitution.However,it is simpler to work directly with the equations (10.26) as in Section 9.7. We verify that these equations can be written as

aijuiuj and A = (a;,) is symmetric, with positive eigenwhere V = values, so that V is positive definite (Problem 4 following Section 10.3). To find the solutions, we set u; = A; sin (At

+E;),

i =l,2

in (10.26). We obtain the equations

and a characteristic equation

The roots are the four distinct real numbers far, f/?, where

and pl = l / m l , pz = I / m 2 (see Problem 1 ) . When A=fa, Eqs. (10.29) are satisfied by A, = k2, A2 = 2k2 - m la2;when A = fB , they are satisfied by A = k 2 , A2 = 2k2 - m 1 g 2Accordingly, the general solution is a superposition of normal modes:

+ c2k2 sin ( p t + C Z ) , u2 = c1(2k2- mlcu2)sin ( a t + < I ) + c2(2k2 - rnlg2)sin (gr + € 2 ) . ul = c l k 2sin (at + € 1 )

The frequencies a,#are I called resonant frequencies.

(10.32)

Chapter 10 Partial Differential Equations

c ) Exponential decay. We assume m 1 = m2 = 0, Fl r 0, F2 E 0, uo = ~3 = 0, h l > 0, h2 > 0. The Eqs. (10.25)become

We seek solutions:

and proceed as in Section 9.6. The characteristic equation is of second degree and has distinct real roots - a , -b, which are both negative (Problem 5 following Section 10.3): A

where q l = I / h , , q2 = l / h2. The general solution is found (Problem 5) to be

We can say: The general motion of the system is an exponential approach to the equilibrium state: ul = u;, = 0.Both "normal modes" are transients for this case, but we can consider the general motion as a linear combination of two such normal modes. The physical model of Fig. 10.1 is not very appropriate here, since we are considering the limiting case ml = 0, m 2 = 0. A better model is suggested in Fig. 10.3. Here we consider a rod made of four pieces, each of which is a perfect conductor of heat, so that the temperature u is constant throughout each piece. The two end pieces are maintained at temperatures uo and u3. It is assumed that heat can flow only across the faces of adjacent pieces according to Newton's Law, so that the rate of flow is proportional to the temperature difference. If h l and h2 denote total specific heats for the middle pieces, we obtain differential equations

Figure 10.3

Model for heat conduction.

665

666

Advanced Calculus, Fifth Edition If kl = k2 = k3 and u3 = uo = 0,we obtain (10.33). It is physically clear that when the end temperatures are maintained at 0, the temperatures of the inner pieces will also gradually approach 0. d ) Equilibrium. We assume that uo and us have constant values u: and u; and that constant forces F, and F2 are applied. Equations (10.25) then have an equilibrium solution, obtained by setting all derivatives with respect to t equal to 0:

-

+

-k2(u2 - 2 ~ 1 u:) = FI,

+ U I ) = F2.

* -2 ~ 2 - k 2 (u3

Accordingly, the equilibrium values of u u2 are

.

(10.38)

.

- u;, w2 = u2 - u;, where u; and u; are defined by (10.39).Then substitution in (10.25),with uo = u;, u3 = u f , Fl = const, F2 = const, yields the equations

e) Approach to equilibrium. Let wl = ul

d2w1 dwl +hl- k (w2-2wl)=0, m1dt2 dt d2w2

dw2 dt2 h2- k (-2w2 + w I )= 0. m2 -

+

dt

(10.40) ,

,~

If, for example, m 1 = m2 = 0 , then these are the same as (10.33), so that

Accordingly,

Just as for one particle, the solution is an exponential appmach to equilibrium. f ) Forced motion. External forces can be applied both by varying the wall positions uo, u3 and through the forces FI , F2. We allow for both effects but neglect the masses by considering the equations

Chapter 10 Partial Differential Equations

.

We now use the method of variation of parameters; we replace the constants C1, C2 in the solutions (10.36) of the homogeneous equations by variables vl(t)? v2(t): i

If we substitute in (10.43) and use the fact that (10.44) satisfies (10.33) when

- the v's are treated as constants, then we obtain equations

for vi , v;. The solutions are vl

=

/

eu'[(2k2 - h l b ) G I - G 2 J d t ,

Substitution in (10.44) yields a solution uT(t), u;(t). The general solution is then

The terms in e-"', emb' can be regarded as transients. The solutions can be interpreted as following an input, defined by (10.39) with u;, u;, FI,F2 replaced by the given functions o f t ; the speed of follow-up depends on the size of h I , h2.

PROBLEMS 1. Show that the roots of (10.30) are given by f a , fp, wherea and @ are defined by (10.31). Show that a and are real and that a! > /I> 0. 2. Solve (10.27) as in Section 9.6 by setting ul = ~ l e ~u2' ,= ~ ~ ew l ~= '~ ,l e " .w2 = ~ 2 e " .Show that (10.23) is again obtained. 3. Let ml = m2 = 1 and k2 = 1, in appropriate units, in (10.26). a) Write out the general solution. b) Obtain the particular solution for which ul =u2 = O for t = O and dulldt = 1, du2/dt = 0 for t = 0. Graph u 1 and u2 as functions of t. Also plot the curve u 1 = u (t), u2 = u2(t) in the u 1 u2 plane; this is a "Lissajous figure." 4. a) Show that Eqs. (10.26)can be written in the form (10.28) with V = k2(ut- ul u2 +u:) and that V is positive definite.

667

668

Advanced Calculus, Fifth Edition .. b) Prove that the expression E = ; r n l ( d ~ l / d t ) ~; r n 2 ( d u ~ / d t ) ~V(u1,uz) is constant for each solution of (10.26).[Hint: Differentiate E with respect to t and use (10.26).] The first two terms give the total kinetic energy; the third term is the potential energy. E is the total energy and remains constant (Section 5.15). 5. Obtain the general solution (10.36)of (10.33)and verify that 0 < b < a . 6. Let h I = h2 = 1, k2 = 1 , in appropriate units, in (10.33). a ) Obtain the general solution. 3D rswfl b) Obtain the particular solution for which ul = 1, u2 = 3 when t = 0 . Graph ul and u2 as functions oft. Also plot the curve ul = ul (t), u2 = u2(t) in the u I u2 plane. 7. Let h l = h2 = 1 , k2 = 1 , Fl ( t ) = 0 , F2(t) = 0 , in appropriate units in (10.43).Let further u0(t) = sint and u s ( ? )G 0 . Obtain the particular solution for which ul = u2 = 0 for t = 0 . Plot u and 2.42 as functions oft and also graph the curve u 1 = ul ( t ) ,u2 = u2(t) in the u 1 u2-plane. 8. Let ml = m2 = 1 , h l = h2 = 0 , k2 = 1 , uo(t) = u 3 ( t )= 0 , F , ( t ) = 4sint, F2(t) = 4a sin t in (10.2.5). Find a particular solution. Does resonance occur? [Hint: use (10.27) and variation of parameters.] 9. a ) Let ml = m2 = 4 , h l = h2 = 1, k2 = 1 , uo = uj = 0, F l ( t ) r F2(t) r 0 in (10.25). Show that the solutions represent damped vibrations. b) Can the values of h h2 be modified so that one normal mode is undercritically damped, while the other mode is overcritically damped?

+

+

.

We now consider the general case of N particles P I , . . . , PNof masses rn I , . . , r n ~ moving on the x-axis as in Fig. 10.4. The particle P, is attached to the particles and Po+,by springs; the particle PI is attached to a wall at Po and to P2; the particle PNis attached to P N - l and to a wall at PN+1.In general, x, is the x-coordinate of P,. For simplicity we assume that all springs have the same natural length 1 and spring constant k 2 .We assume that P, is subject to a resistance -h, and an outside force F,(t). The differential equations corresponding to (10.21) are then the following:

9

and F,(t) = 0 for a = 1, . . . , N, then If the walls are fixed: xo = x,*,xN+1 = there is an equilibrium state, determined by the N equations: ~,+~-2x,+x,-l=O,

a = 1 , . . . , N.

Equations (10.48) can be written as follows:

0 x 0

XI

Figure 10.4

X2

I

Xa

(10.48) rr

%+I

Linear system with N particles.

Chapter 10 Partial Differential Equations

Figure 10.5

N particle model for the vibrating string.

They assert that P, is halfway between and P,+I. Hence at equilibrium, all particles are equally spaced between xo and xN+l:

We now refer the motion of the particles to the equilibrium positions (10.49) by introducing new coordinates: u,=x,-x:

t\

( a = O,...,N + l ) .

*?

2

(10.50) fl,

$1

The differential equations (10.47) are then replaced by the following:

~iil

A second physical model that leads to Eqs. (10.5 1) is suggested in Fig. 10.5. Here the particles P I , . . . , PN are constrained to move along lines x = X I , . . . ,x = x~ ,,, in the xu-plane. Again P, is attached to Po+]and to Po-] by springs; Po and PN+1 are points on the "walls": x = xo, x = X N + We ~ . assume that the lines x = x, are equally spaced, at distance Ax and that all particles have the same spring constant k,Z and natural length 1, where 1 < Ax. If P, has coordinates (x,, u,) and is subject to a resistance -h,(du, l d t ) and an outside force F,(t), then the differential equations governing the motion are as follows:

d2u, du = k2O(U,+I - 2u, m, -+ h, -2 dt'

dt

+~

-

~ - 1 )k$(sincu,

- sins,-,) + F,(t),

-

gm+].

where a = 1. . . . , N and a, is the angle from the positive x-direction to If the angles cr, remain so small that an approximation sin a, tan a, is justified, the equations become

where k2 = k;[l - ( / / A x ) ] .The derivation of (10.52) and (10.53) is left to Problem 1 following Section 10.4. Equations (10.53) are identical with (10.51). When m, = 0 for a = 1, . . . , N, a natural model can be devised by generalizing the heat conduction model of Fig. 10.3. Other models can be constructed, for example, by use of electric circuits. '

I

f

>

670

Advanced Calculus, Fifth Edition'

-

Equations (10.51) can be written in a form that suggests further generalizations. We write V is the potential energy associated with the system. Equations (10.51) then become

ex'Cept for a = 1 and a = N ;the exceptional cases can be included if we modify the definitions of Fl(t) and FN(t)to include k2uo(t)and k ' u ~ + ~ ( trespectively. ), If we allow V to be a general function of u l , . . . , UN, instead of the special function (10.54), then an extremely broad class of physical problems is included in (10.55). Of great importance is the case in which V is a general quadratic expression:

As observed in Section 9.7, this case arises in consideration of problems of "small vibrations," that is, problems concerning the vibrations of a system of particles that remain close to equilibrium positions. The problems of Figs. 10.4 and 10.5 are of this sort. The approximation sin a -- tana was based on the smallness of the departure from equilibrium. We now state briefly the results corresponding to (a), . . . , (f) for (10.51). Actually, many of the statements apply equally well to Eqs. (lo.%), where V is defined by (10.56) and is positive definite. The positive definiteness of V in (10.54) is proved in Problem 2.

a) Harmonic motion. Here h , = 0, uo = U N + I = 0, F,(t) = 0. We find that there are N normal modes of vibration, with resonant frequencies h 1 , . . . , AN, and that the general motion consists of a linear combination of these: N U,

=

c , ~ , , , sin (Ant

+ en);

(10.57)

n=l

the A,,, are certain constants, and e l , . . . , CN , t I , . . . , C N are arbitrary constants. * b) Damped vibrations. In this case, m, > 0, h, > 0,uo = U N + ~= 0, F,(t) 3 0. The oscillations of (a) are replaced by damped oscillations, of form e-"' sin(At c), or exponential decay terms: e-"', re-"'. c) Exponential decay. Here m, = 0, F, = 0, uo = U N + ~ = 0, h, > 0. We find that there are N "normal modes of decay" and that the general motion is a linear combination of the normal modes: (EE

+

N

u, =

f rrl

c.~,,,e-~~'. Ii (10.58) n=l d) Equilibrium. If uo = u:, u N + ~= u*,+,,F, = F,*,where the starred values are constants, then the equilibrium values u, = u: are equally spaced ( ~ ; " / k ' )and u*,+, ( ~ i / k ~ ) . between u:

+

+

Chapter 10 Partial Differential Equations

e) Approach to equilibrium. Under the assumptions of case (d) the general motion u,(t) is a solution of the homogeneous equations corresponding to (10.5 1) plus the equilibrium solution u:. When the homogeneous equations are of type (c), one therefore has the exponential approach to equilibrium: N

u, = ~ c n ~ n , , e - a t z ' + u(o ~ = 1. . . . , N). n=l

(10.59)

f ) Forced motion. Here uo, U N + I and , all F, are allowed to depend on t. The

,

general motion u, ( t )is a solution of the homogeneous equations corresponding to (10.51) plus a particular solution u:(t). The particular solution can always be found by variation of parameters; it can always be interpreted as a follow-up of an input. The total effect of all F,(t) on the solution can be considered as a superposition of the effects of the F,(t) individually. Addition of a sinusoidal term of frequency h to one F, leads to the addition of a sinusoidal term of frequency h to all u:(t), unless all h, are 0 and h is a resonant frequency, in which case resonance may occur (cf. Problem 8 following Section 10.3).

PROBLEMS 1. a) Obtain the Eqs. (10.52) for the model of Fig. 10.5 b) Show that when the angles cr, remain small, the Eqs. (10.53) are justified as approximations to (10.52). 2. a) Show that the potential energy V of (10.54) is positive definite. [Hint: Write V = (k2/2)[ui ( U I- ~ 2 ). . ~ . U~I. b) Show that when h, = 0 and F,(t) = 0 for a = 1, . . . , N, every solution of (10.55) has the property that the total energy 1 1 E = -ml ( d u ~ l d t ) ~ - - i m ( d~u ~ l d t ) ' V(ul, . . . U N ) 2 ,\' is constant. [Hint: See Problem 4 following Section 10.3.1 3. Let N=3in(10.51)andletm~=m~=m3=l,k2=1,hl=h2=h3=O,uO=u4=0, F1= F2= F3 = 0, so that one has case (a). Seek solutions: u I = Al eA', u2 = A2eU', u3 = A3e" and obtain the general solution in the form (10.57). 4. Let N=3in(10.51)andletml = m 2 = m 3 = 0 , k 2 = l , h l = h 2 = h 3 = 1 , u o = u 4 = 0 , FI = Fz = F3=0, so that one has case (c). Seek solutions: u1 = ~ ~ e u2 " ,= A2eA', u3 = A3eA1and obtain the general solution in the form (10.58). 5. (D~fferenceequations) We consider functions f ( a ) of an integer variable a : a = 0, f1, f2, . . . . Let f ( a ) be defined for a = m, a = m + 1, . . . , a = n. Then the first difference A+ f ( a ) is the function f ( a 1) - f ( a ) (m 5 a in); the first difference A - f ( a ) is the function f ( a ) - f ( a - 1 ) (m < a 5 n). The second difference is the first difference of the first difference; this could mean A+ A- f , A- A+ f , A+ A+ f , A- A- f ; we shall use only

+

+ +

+ +

+

.

+

see Problem 9 following Section 6.18. Differences of h~gherorder can be defined in analogous fashi~q.A difference equation is an ident~tyto be satisfied by f (a),and its

671

672

Advanced Calculus, Fifth Edition differences of various orders. In general, the "linear difference equation" has a theory quite analogous to that of the linear differential equation, and the functions era play a similar role. We consider only two cases: a) a2f ( a ) = 0. Show that the general solution is given by f ( a ) = c l a c2, where cl, c2 are constants. Obtain the solution satisfying the boundary conditions: f (0) = uo, f (N 1) = U N + where ~, uo. u ~ + are l given constants. f ( a ) p 2f ( a ) = 0. Show that the functions cl cos q a + c2 sin q a are solutions, b) provided that cos q = 1 and 0 < p2 < 4. Show that when p 2 > 4, distinct numbers a , and a2 can be chosen so that the functions claF + c2a; are solutions and cza(- l), are solutions. that when p2 = 4, the functions c l (- l), c) Show that if f (0) and f (1) are given, then the difference equation of part (b) successively determines f (2), f (3), . . . . Hence for such "initial conditions" there is one and only one solution. Determine the constants ci , c2 for each of the cases of part (b) to match the given initial conditions. The fact that this is possible ensures that each expression gives the general solution for the corresponding value of p. d) Show that the only solutions of the difference equation of part (b) which satisfy the boundary conditions: f (0) = 0, f (N 1) = 0 are constant multiples of the N functions

+

+ +

kp2

+

+

and that @,,(a) is a solution only when p 2 = Z(L

).

- cos N ""+I

6. Show that the solution of (10.48), where xo = x:, X N + ~ = x ; J + ~is equivalent to solution of the difference equation f ( a ) = 0 of Problem 5(a), subject to boundary conditions, and compare with the solution of Problem 5(a). 7. Let rnl =m2= ... = m ~ = r n , h , = O and F,(t)=O for a = 1 , ..., N and u o = U N + I = 0 in (10.51). so that one has case (a), with equal masses. Show that the substitution u, = A(n) sin (At + E ) leads to the difference equation with boundary conditions:

Use the result of Problem 5(d) to obtain the N normal modes u,(t) = sin

(E -

a) sin (Ant

+ s),

ShowthatO
+ 1, we define an inner

673

Chapter 10 Partial Differential Equations the norm 1 1 f I I is then defined as ( f , f )f. In the following we consider only functions that equal 0 for a = 0 and a = N + I . In particular, we use the functions & ( a ) of Problem 5(d):

a) Graph the functions $,(a) for the case N = 5. b) Show that (@,,

& ) = 0 for m # n and that 114,112 = ;(N

+ 1). [Hint: Write

and evaluate inner product and norm with the aid of the formula for sum of a geometric progression.] C) Show that if we associate with each function f ( a ) the vector v = [ v l, v2, . . . , v N ] , where vl = f ( 1 ) , v2 = f ( 2 ) ,. . . , U N = f ( N ) , then the operations f g, c f , ( f , g) correspond to the vector operations u + v, cu, u . v of Section 1.14. Accordingly, the space of functions considered forms an N-dimensional Euclidean vector space. The form a system of base vectors. vectors corresponding to the functions 4,(a)/I 14n(a)() d) Show that if f ( a ) is defined for a = 0 , . . . , N 1 and f ( 0 )= f ( N 1) = 0, then f ( a ) can be represented in one and onIy one way as a linear combination of the functions +n(a),namely, as follows:

+

+

+

Compare with the Fourier sine series (Section 7.5). 9. Write the general solution of Problem 7 in the form

where anand Bn are arbitrary constants. Use the result of Problem 8(d) to show that the constants a, and /In can be chosen in one and only one way so that u,(t) satisfies given initial conditions:

This shows that one has indeed obtained all solutions. 1 0 . L e t h I = h 2 = ~ ~ ~ = h N = h , m n = O a n d F , ( t ) ~ O f o r a, =. .1. , N , u o = u ~ + l = 0 in (10.51), so that one has case (c), with equal friction coejicients. Show that the substitut~onu, = A(a)eArleads to the difference equation with boundary conditions:

+

A ~ A ( U ) p 2 ~ ( a=) 0 , 'tlP,

A(0) = 0 ,

p2 = - h h / k 2 , A(N

+ 1 ) = 0.

Use the result of Problem 5(d) to obtain the "modes of decay":

u b ( t ) = sin ( $ a )

e-atnt.

-I

Advanced Calculus, Fifth Edition .)!of . 11. Prove that constants c, can be chosen in one and only one way so that

is a solution of the exponential decay problem (Problem 10) and matches given initial conditions: u,(O) = f (a) (cf. Problem 9). 12. In (10.51), let h l = h z = ... =h,m,=O, f o r a = l , ..., N as in Problem 10, but let F,(t) be permitted to depend on t. We assume uo = 0, U N + ~= 0, as any variation in the "walls" can be absorbed in Fo(r) and FN+(t).Use the method of variation of parameters to obtain a particular solution. [Hint: The "complementary function" is given in Problem 11. If we replace cn by vn(t) for n = 1, . . . , N and substitute in (10.51). we obtain equations

Now use the result of Problem 8(d) to conclude that

We now consider the limiting case: N -+ oo. Rather than attempt to carry out a precise passage to the limit, we allow ourselves to be guided by physical intuition. The natural limiting case of the system of N particles moving on a line (Fig. 10.4) is that of a rod that is permitted to vibrate longitudinally, as suggested in Fig. 10.6. The individual particle is replaced by a cross section of the rod, which can be thought of as a thin layer of molecules that move together parallel to the axis of the rod. When no external forces are applied, this layer has an equilibrium position x. Just as for the particles, we can measure the displacement u of the layer from its equilibrium position x; u then becomes a function of x and t. If we pass to the limit in the model of Fig. 10.5, we obtain the vibrating stringfor example, a violin string. To first approximation, each "molecule" of such a string executes vibrations perpendicular to the line represented by the equilibrium position of the string. The displacement of the molecule at position x from its equilibrium position is measured by u, which is a function of x and t. The vibrations are assumed to take place in an xu-plane; one could consider the more general case in which the vibrations are not confined to a plane.

Figure 10.6

Longitudinal vibrations of a rod.

.,

Chapter 10 Partial Differential Equations In order to obtain a differential equation for u(x, t ) we return to the basic equations (10.51) and write them as follows:

+

We assume .yo and XN+, to be fixed and let L = X N + ~- x0, Ax = L / ( N 1). We then let N increase indefinitely. The ratio m , / A x represents an "average density" at position x ; it is reasonable to postulate that this approaches as limit a function p ( x ) representing density (mass per unit length) at x . The simplest law of friction would make h , proportional to m,, so that h , / A x would approach a function H ( x ) of dimensions force per unit of length per unit of velocity. For the model of Fig. 10.4 the product k 2 a x represents the tension in one of the springs when it is stretched a distance Ax. But precisely the same tension must hold in each half of the spring, which is stretched only a distance : A X . Hence if we always use springs of the same stiffness, k 2 a x will approach as limit a constant force K'. For the model of Fig. 10.5, one indeed has k 2 a x = k i ( ~ x I ) , where ki is the actual spring constant for each spring; therefore k 2 a x represents precisely the tension in each spring when all displacements u, are 0 ;the limiting value K' is precisely the tension in the string. One can write

where x, is the equilibrium position of P,. In the limit, x, becomes a continuous variable, and, as in Problem 9 following Section 6.18, the ratio of the second difference to (AX)' approaches as "limit" the derivative

The right-hand members we assume to approach as limit a function F ( x , t ) representing applied force per unit length at x . We are thus led to the partial differential equation

a 2 +~ H(x)-au

p(x)-

,a2u

- K- - = F ( x ,t ) ,

at2 at ax2 with p(x) 2 0 , H ( x ) 1 0 , K > 0. This is the fundamental partial differential equation to be studied. Certain generalizations will be introduced in later sections, notably by replacement of a2u/ax2by the Laplacian V'u :

p-

aZu + H -au at2

at

-K ~ V = ~ U F(x,y , Z, t ) .

This corresponds to a generalization to motion in 2- or 3-dimensional space. One can easily construct an N particle model for this. From the general equation (10.55), one obtains a broader class of equations in which the term - K ~ V is ~ replaced U by a more complicated expression, possibly nonlinear, in u and its derivatives. Two- and three-dimensional problems can lead to simultaneous partial digerential equations. While the generalizations do introduce complications, the principal problems and methods reveal. @emselves in Eq. (10.61) and, in fact, in the N particle

676

Advanced Calculus, Fifth Edition approximation to this of the preceding section; as was pointed out in the introductory section, even the single particle displays the properties that are crucial. The limiting process by which we arrived at (10.61) was based on physical intuition, and the basic test of the validity of the result is its accuracy in explaining the behavior of continuous media. This is a problem of physics, by no means simple, with which we shall not concern ourselves. However, one can ask the purely mathematical question: Do the solutions of difference equations converge to the solutions of the corresponding differential equations, when the basic interval (for example, Ax) approaches O? This question has been made precise and answered in a generally affirmative fashion in recent research. We refer to the book by Isaacson and Keller and that of Iserles listed at the end of the chapter for a discussion of the problem and further references to the literature.

Equations (10.61) and (10.62) are linear in u and its derivatives and are hence linear partial differential equations. They involve derivatives of u up to the second order and are hence partial differential equations of second order. The most general linear partial differential equation of second order in two independent variables has the form

where A , . . . , G are functions of x and y. Equations (10.63) are classified into three types:

+ 2 B t q + c q 2 = 1 is an ellipse; B2 - AC = 0, A t 2 + 2 B t q + cV2 + D t + Eq = 0 is a parabola; B2 - AC > 0 , ~5~ + 2BCq + c q 2 = 1 is a hyperbola.

elliptic: B* - AC < 0, A t 2 parabolic: hyperbolic:

An equation can be of one type in part of the xy-plane and of another type in a second part. An analogous classification is made for equations in three or more independent variables. The three types are illustrated respectively by the equations

The first of these is the equation V 2 u = 0 and occurs naturally in connection with the equilibrium problem for two dimensions. The second corresponds to exponential decay; the third to harmonic motion. The differential equation (10.61) was proposed as a natural one for the longitudinal oscillations of a rod, as in Fig. 10.6, or for the transverse vibrations of a string. There are other 1-dimensional problems to which the equation is applicable: Planar sound waves and electromagnetic waves, diffusion of heat, and other diffusion processes ( p = 0).The equation can also be applied to inJinite intervals on the x-axis; although the vibration of a string infinite in length may appear to be an artificial

Chapter 10 Partial Differential Equations concept, such an ideal case is useful in applications. Equation (10.62) has analogous applications in two and three dimensions, including the basic hyperbolic, parabolic, and elliptic equations: , wave equation: heat equation: Lnplace equation:

v2u = 0;

here p , H, and K2 are usually considered to be positive constants. As was pointed out in Section 5.15, the Laplace equation is satisfied by the velocity potential of an irrotational, incompressible fluid motion. The complete equations of hydrodynamics are simultaneous equations that are nonlinear (see Lamb's Hydrodynamics, Dover Publications, 1945). The basic problems associated with (10.61) are simply the analogues for the continuous medium of the problems studied in the previous sections. For example, problem (a) concerns the case for which p(x) > 0, H (x) = 0, F(x , t) = 0, and the "walls" are fixed: u(0, t) = 0, u(L, t) = 0; we expect to show that there is one and only one solution to the initial value problem: u = f (x), aulat = g(x) for t = 0. Such a solution u(x, t) would be defined and continuous for 0 < x 5 L and for t 2 0 and would be required to have partial derivatives through the second order for 0 < x < L and t > 0 and to satisfy the differential equation in the domain just described. One can also consider the possibility of discontinuities on the boundary; this requires care, but significant results can be obtained. The word solution will refer to functions u(x, t) continuous for 0 5 x 5 L , t > 0, unless otherwise indicated. Problems (b) and (c) (damped vibrations and exponential decay) are formulated in a similar manner. The equilibrium probkrp (d) now becom9s.w ordinary diffqeptial equation: ,,

with boundary conditions: u = uo for x = 0, u = ul for x = L. In two dimension? k the analogous equation is the Poisson equation -K2v2u = F(x, y), where u has given values on the boundary of a 2-dimensional region; when F = 0, this is the Dirichlet problem. In all cases we wish to show that there is a unique equilibrium state u*(x) [in two dimensions, u*(x, y)]. As for the N particle problem, the approach to equilibrium, problem (e), is described by a function u*(x) u(x, t) where u(x, t) is a solution of a homogeneous problem (a), (b), or (c). The problem (f) of forced motion includes the other five problems as special cases. Boundary conditions: u(0, t) = uo(t), u(L, t) = ul(t) and initial values of u are given; we wish to show that there is one and only one corresponding solution. The methods to be used are a natural extension of those used for the N particle problem. The homogeneous problems are handled by a substitution: u(x, t) = A ( x ) d t , which gives the normal modes; the "general solution" is again obtained

+

677

678

Advanced Calculus, Fifth Edition as a linear combination of normal modes. The nonhomogeneous problem of forced motion is solved by variation of parameters; the general solution is the sum of a particular solution u*(.r, t ) and the general solution of the homogeneous problem. We have considered only the cases of fixed walls or walls moving in a prescribed manner. There are other natural boundary conditions; for example, one could require that au/ax be 0 for x = 0. For the N particle case this would correspond to the requirement that the wall Po move in such a fashion that uo = ul ; accordingly, the distance between Po and PI is fixed, and no energy can be transmitted. For the heat conduction problem this corresponds to an insulated boundary at x = 0.

In the basic equation (10.61) we assume that H(x) = 0 and F(x, t) r 0 and that p is constant, independent of x. The differential equation becomes

The equation is to be applied to a rod or string occupying the po';tion of the x-axis between x = 0 and x = L. We assume that the ends are fixed:

By a change of scale, x' = n x / L , we can reduce the problem to the case for which L = IT.The equation becomes

For simplicity we drop the prime in the following. We introduce the abbreviation:

The equation and boundary conditions now read

To determine the normal modes, we can now set

in (10.66) and (10.67). However, we find as in the N particle case that A would have to be pure imaginary (Problem 5 ) , and we simplify the process by replacing (10.68) by the substitution U(X)= A(x) sin (At + 6).

(10.68')

,

Chapter 10 Partial Differential Equations Equations (10.66) and (10.67) then become the two equations

The simultaneous linear equations of the N particle problem are therefore replaced by a differential equation with boundary conditions. This is anticipated in Problems 5 to 12 following Section 10.4, in which it is shown that the simultaneous equations of the N particle problem can be treated as a difierence equation with boundary conditions. The solutions obtained there are very closely related to those to be obtained for (10.69) and (10.70). nn The general solution of (10.69) is

):(

~ ( x=) cl sin

+ c2cos ();.

Equations (10.70) are satisfied only if

One obtains the characteristic values (resonant frequencies or eigenvalues)

and associated characteristic functions (or eigenfunctions) An(x) = sin nx

(n = 1,2, . . .).

>

.

"

(10.73)

We restrict to positive A, since a change in sign can be absorbed in the phase constant r . The normal modes are

-

~31

(10.74)

sin nx sin (ant -ken)

and constant multiples thereof; there are infinitely many normal modes. The set of frequencies A, occurring is called the spectrum. We now attempt to construct the general solution u(x, t) as a linear combination of normal modes: * I

L

CO

U(X,

t) =

1cnsin nx sin (ant + 6.). n=l

However, we face a new difficulty: The injinite series (10.75) may fail to converge. Even if it does converge, it may fail to satisfy the differential equation (10.66), for this requires the existence of second derivatives. Now the series (10.75) can be regarded as a Fourier sine series in x , with coefficients dependent on t. From the theory of Fourier series (Chapter 7) we easily obtain conditions on the constants cn such that the series converge for all x and can be differentiated twice with respect to x and t.

679

680

Advanced Calculus, Fifth Edition The choice of the constants in (10.75) depends on the initial conditions, for we can write (10.75) in the form: U(X,t) =

n=l

sinnx[an sin(ant) + #?, cos (ant)],

Bn = Cn sin en.

an = Cn COS En,

(10.75')

Then, if we assume that the series concerned are uniformly convergent, 00

u(x, 0) =

C Pn sinnx,

n=l

au

QO

- (x. 0)

at

=

n=l

naan sin nx.

(10.76)

Thus fin and naa, are the Fourier sine coefficients of the initial displacement and velocity, respectively. THEOREM

If the constants c, are such that cnn4is bounded:

then the series (10.75) converges uniformly for all x and t and defines a solution of the wave equation (10.66) for all x and t. Let f(x) and g(x) be defined for 0 ( x 5 IT;let f (x) have continuous derivatives through the fourth order and let f (0) = f (n) = f "(0) = f "(n) = 0; let g(x) have continuous derivatives through the third order and let g(0) = g(n) = gN(0) = gf'(n) = 0. Then there exists a solution u(x, t) of the wave equation (10.66) with boundary conditions ( 1 0.67) such that

namely, the series (10.75'), where

1"

2 = g(x) sin nx dx. (10.79) an = i n o nan The solution is unique; that is, if u(x, t) satisfies (10.66), (10.67), and (10.78) and the partial derivatives u,, , u,, are continuous for 0 5 x _( n , t 1 0, then u(x, t) is necessarily represented by the series (10.75'), with coefficients given by (10.79).

pn = - J f(x)sinnx dx,

Proof. If (10.77) holds, then the Weierstrass M-test (Section 6.13) shows that the series (10.75) converges uniformly for all x and t. Similarly, the series

-

n2cnsin nx sin (ant

+ en),

-

1n2a2cnsin nx sin (ant + 6)

obtained by differentiating (10.75) twice with respect to x and t converge uniformly for all x and t; for by (10.77) ln2cnI < Mn-2. Accordingly, these series represent u,, and u,,, respectively. By substitution in (10.66) we verify that the wave equation is satisfied. Iff (x) and g(x) satisfy the conditions stated, then (Problem 4 following Section 10.8) n4anand n4Pn are bounded so that

681

Chapter 10 Partial Differential Equations is bounded. Accordingly, (10.77) holds, so that (10.75) or (10.75') represents a solution u(x, t) that is continuous for all x and t. When t = 0, the series for u and u, reduce to the Fourier sine series of f (x) and g(x); these series converge to f (x) and g(x) (Problem 1 following Section 7.13). The proof of uniqueness is left to Problem 6 following Section 10.8.

10.8 PROPERTIES OF SOLUTIONS OF THE WAVEEQUATION We consider the solutions in the form 00

U(X,t ) =

sinnx[a, sin (nat) + B. cos (nat)].

(10.80)

n=l

For each fixed x the series is a Fourier series in t, with period 2 n l a (Section 7.5). Accordingly, each point of the rod or string considered moves in a periodic fashion, with period 2nla. The normal mode for which n = 1 is called the fundamental mode. Here U(X,t) = sinx [alsin (at)

+ fiI cos (at)].

This is easily visualized in the case of a vibrating string, for which the displacement has the shape of a sine curve at all times (Fig. 10.7). The string is therefore vibrating in this shape with frequency a/2n (cycles per unit time). The modes corresponding to n = 2,3, . . . are called thejrst overtone, second overtone, and so on; musically, they give the octave, octave plus a fifth, and so on. These are suggested in Fig. 10.7. They are easily demonstrated on a stringed instrument, especially on low notes such as low C on a violoncello. We remark that the shapes of the normal modes are precisely the characteristic functions A,(x) = sin nx and that these functions form a complete orthogonal system for the interval 0 5 x 5 n. The relation between solution and initial conditions can be shown in a striking way by the following observation. Let us first assume that g(x) = 0, so that the rod (or vibrating string) is initially at rest but has an initial displacement f (x). By (10.79), a, = 0 for all n. We can now write

xBn 00

U(X,t) =

n=l

lc0 sinnx cosnat = - x B n [ s i n n ( x + a t ) 2 n=l

+ sinn (x -at)].

Since 00

f (x) =

fin sin nx

.

(10.81)

n=l

we can write: 1

U(X,

Figure 10.7

t) = -[ f (X+ a t ) 2

+ f (x - at)].

Normal modes for vibrating string.

(10.82)

.)

682

Advanced Calculus, Fifth Edition

Figure 10.8

f ( x ) versus f ( x

+at),f (x - at).

Figure 10.9 Solution u ( x , t ) of the wave equation. The curves show the wave forms for t = 0, 1 , . . . , 10. The units are chosen so that the wave velocity a is 0.2 L per unit of time.

.

This representation is at first valid only for

However, if we extend the definition of f ( x ) to all x by (10.81), then (10.82) has meaning for all x and t and, under the assumptions of the theorem above, represents a solution of the wave equation for all x and t. The term f (x + a t ) represents the initial displacement translated at units to the left; the second term represents this displacement translated at units to the right; this is suggested in Fig. 10.8. In Fig. 10.9, f ( x ) is chosen as a displacement confined almost entirely to an interval in - 6 < x < in 6, where 6 is small; the solution can then be plotted as a function of x and t. The disturbance is seen to split into two disturbances that travel in opposite directions until they reach the walls, where they are reflected, with a change in sign, and move back together. This can be demonstrated experimentally in various ways-by displacing and releasing a violin string or by sound echoes, for example. If the initial displacement f (x) has a jump discontinuity, for example, at xo, but is piecewise very smooth, then (10.82) continues to define a solution of the wave equation except for x fat = xo fkn. These lines are the paths of "propagation of discontinuities"; they are called characteristics.

+

Chapter 10 Partial Differential Equations The constant a appears as the velocity with which the disturbance or discontinuity is propagated to the left and right; it is termed the wave velocity. If we choose as u a single mode of vibration: 1 U ( X , t ) = -pn[sinn(x + a t ) + sirin(x - a t ) ] , 2 then the initial disturbance f (x) is simply B,, sin n x , a wave repeating itself at intervals of 2rr/n, the wavelength. The oscillations at each x have a frequency of n a / 2 x (cycles per unit time). Hence 2 x na (wavelength) . (frequency) = - . n 2~t = a = wave velocity. This is one of the fundamental rules of physics. We now consider the case in which f (.r) = 0 but the initial velocity g(x) is idifferent from zero. Now, as before, 00

g(x) =

C naa. sin nx.

Hence we can integrate (Section 7.13):

The solution u ( x , t ) is therefore M

u(x, t ) =

1 =-

sinnx sinnat

1

x+at

g(s) ds,

2a I-,, where s is a dummy variable of integration. The representation (10.85) is valid for all x and t if we use (10.83) to extend the definition of g ( x ) to all x . Equation (10.85) can be interpreted as the difference of two disturbances moving to the left and right with velocity a. If we allow for both initial displacement and initial velocity, the same reasoning as before leads to the general formula

Remark. For the wave equation one may consider other natural boundary conditions, as at the end of Section 10.6. The shapes of the normal modes continue to yield complete orthogonal systems. This can be verified in particular cases as in the problems following Section 7.13 or as a consequence of a general property of Sturm-Liouville boundary value problems, discussed in Section 10.13. A similar statement applies to the heat equation (Section 10.9).

683

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Advanced Calculus, Fifth Edition

PROBLEMS

I

.

?

*

1. Let f (x) be an odd periodic function of period 2 n and let f (x) = 0 for 0 < x < 1713 and for 2n/3 < x < n , f (x) = 1 for n / 3 < x < 2x13. Show that (10.82) defines a solution of the wave equation for a certain portion of the xt-plane; analyze the solution graphically as in Figs. 10.8 and 10.9. Is the Fourier series solution valid w~ththis f (x) as the i n ~ t ~ a l displacement? 2. Let g(x) be an odd periodic function of period 2n and let g(x) = 0 for 0 < x < n / 3 and for 2n/3 < x in , g(x) = 1 for n / 3 < x < 2x13. Show that (10.85) defines a solution of the wave equation for a certain portion of the xt-plane; analyze the solution graphically. Is the Fourier series solution valid with this g(x) as the initial velocity? 3. Show that the change of variables r = x a t , s = x - a t , converts the wave equation (10.66) into the equation

+

Interpret the change of variables geometrically. Show that the "general solution" of (a) has the form Discuss the relation between this representation and (10.86). 4. a) Prove that if f (x) satisfies the conditions stated in the theorem of Section 10.7, then p,n4 is hounded. [Hint: Use integration by parts as in Section 7.8.1 b) Prove that if g(x) satisfies the conditions stated in the theorem of Section 10.7, then a n n 4is bounded. 5. Show that the substitution (10.68) in (10.66) and (10.67) leads to equations that are satisfied only when h, = ani and A = An(x) = c(einX- ePinX).From these expressions, obtain the normal modes (10.74). 6. Prove that under the conditions stated in the theorem of Section 10.7 the solution u(x, t) satisfying initial conditions (10.78) must have the form (10.75') and is hence uniquely determined. [Hint: Under the assumptions made, u(x, t) has a representation as a Fourier sine series in x:

Differentiate the second equation twice with respect to t, using Leibnitz's Rule (Section 4.9) and integration by parts to show that d i ( t ) a2n2@,,(t)= 0. Hence 4,(t) = cr, sin (nut) B, cos (nut).] 7. To raise the pitch of a violin string, one can (a) increase the tension, (b) decrease the density, (c) shorten the string. Show how these conclusions follow from (10.72) and (10.65). 8. Find the general solutions of the following partial differential equations for 0 < x < n , t > 0, with boundary conditions u(0. t ) = u(n, t) = 0:

+

+

a)

3at+ + Uax= ~

b) ga2u - 9a 2 u- 2 au~ --

9. Find the general solution of the wave equation

such that u(0, t) = u(2n, t ) , u,(O, t ) = u,(2n, t).

.

Chapter 10 Partial Differential Equations 10. Find the general solution of the wave equation

:-

%

such that u(0, t ) = 0,u,(n,t ) = 0.

We return to the basic equation of Section 10.5:

We neglect masses; that is, we set p = 0. We assume that H is a constant, that there is no outside force F, and that the "walls" are fixed. The differential equation and boundary conditions become

Equation (10.87) is the 1-dimensional heat equation. The 3-dimensional heat equation

was derived in Section 5.15 [Eq. (5.144)]; Eq. (10.87) can be regarded as the special case of heat conduction in an infinite slab bounded by planes x = 0, x = L in space, the boundary conditions being such that the temperature depends only on x. One can also interpret the equation as describing conduction of heat in a thin rod, insulated except at the ends. The transition from a system of masses attached by springs to the problem of heat conduction may at first seem artificial. However, the solutions of (10.87) do indeed display the properties of a limiting case of the system of masses as the friction increases and the total mass approaches 0. Because masses are neglected, disturbances can be propagated instantaneously; the wave velocity is inJinite. As in the preceding sections, we introduce a new variable x' = n x / L and make the substitution:

After dropping the primes the eiuation and boundary conditions become

685

686

Advanced Calculus, Fifth Edition The substitution u =~(x)e"

(10.93)

leads to the characteristic value problem:

If h is positive or zero, the only solution of (10.94) and (10.95) is the trivial solution: A(x) 0 (Problem 9 following Section 10.11). If A is negative, the solutions are the characteristicfunctions An(x) = sinnx

(n = 1,2, . . .),

(10.96)

with associated characteristic values An = -n 2 c2 .

Accordingly, the "normal modes" are the functions

and constant multiples thereof. We expect the general solution to be given by the linear combinations bO

U(X,

t) =

b,, sin nx e-""",

(10.99)

where the bn are arbitrary constants. Again we have to be careful about convergence. The problem is simpler than that for the wave equation, for if the coefficients bn are bounded-lbnl < M for n = 1,2, . . .-then the series (10.99) converges uniformly in .x and t in each half-plane: t 1: t , , -oo c x < oo,provided that t , > 0.This is a consequence of the Weierstrass M-test (Section 6.13), for

Ib, sin n x e-'2n2tI < M ~ - c ? I I=~ ~M,; , the series M, is a series of constants whose convergence follows from the ratio test. In the same way we verify that the series (10.99) remains uniformly convergent in the range stated, when differentiated as often as desired with respect to x and t (Problem 5 following Section 10.11). In particular, . .i4

so that ( 10.99) is a solution of the heat equation in the range t > 0, -00 < x < CQ. The boundary conditions (10.92) are satisfied, since each term in the series is 0 when x = 0 or x = n. Every sufficiently smooth solution of the heat equation and boundary conditions in the range described must have the form (10.99) (Problem 7 following Section 10.11).

687

Chapter 10 Partial Differential Equations For t = 0 the series (10.99), if convergent, reduces to u(x, 0)=

bn sinnx. n=l

Thus, just as for the wave equation, the initial values of u, u(x, 0) = f (x), are represented by a Fourier sine series. If, for example, f (0) = f (n) = 0 and f (x) has a continuous second derivative for 0 5 x 5 n , then its sine series will converge uniformly to f (x), and the series (10.99) will converge uniformly for 0 5 x 5 n and t 2 0. Accordingly, under the assumptions stated, the series (10.99) defines a function u(x, t) that is continuous for t 2 0,O 5 x 5 n and that satisfies the heat equation (10.91), the boundary conditions (10.92), and the initial condition u(x, 0) = f (x). The solution is furthermore unique (Problem 7 following Section 10.11). These results show that with minor modifications the theorem of Section 10.7 holds for the heat equation. Iff (x) is merely piecewise continuous, the constants I

l

. i r

A-

'

-

bn =

1Jr

f(x)sinnx dx

(10.101)

0

are bounded, since the sequence bn converges to 0 [Eq. (7.14), Section 7.41. Accordingly, the series (10.99) defines a solution of the heat equation for t > 0. The series i converges to f (x) in the mean (Section 7.1 1) for t = 0; in fact, it can be shown that, lim u ( x , t ) = f(x),

0 < X< n

t+o+

for each x at which f (x) is continuous. (See pages 279-290 of the book by Doetsch listed at the end of the chapter.)

10.10 PROPERTIES OF SOLUTIONS OF THE HEATEQUATION For each fixed x, each term of the series (10.99) describes an exponential approach to 0. A similar statement applies to the sum of the series; that is,

-

limu(x,t)=O t-+m

(Problem 6 following Section 10.11). The rate of decay varies with n; the highfrequency terms in (10.99) have smaller exponential coefficients than the lowerfrequent) terms and are accordingly damped out more rapidly. This corresponds to the observed fact that abrupt variations in temperature in an object disappear rapidly, whereas differences in temperature at large distances are evened out slowly. If we write ( n ~ ) ~ ( n ~ ) ~ sinnx = nx - -+--... 3! 5! in (10.99) and collect terms in powers of x, we obtain a series:

688

Advanced Calculus, Fifth Edition This operation can be justified by theorems on series or, more simply, by complex variables as in Problem 4 following Section 10.11. Accordingly, for each fixed t > 0 the solution u ( x , t ) can be represented by a power series in x ; the power series has an infinite convergence radius, so that the series converges for all x. The solutions u ( x , t ) are analytic in x. From this last result we deduce another property of the solutions-the infinite speed of propagation of disturbances. For example, let the initial function f ( x )be 0 for 0 5 .Y 5 rr and greater than 0 for xl < x < x2, where 0 < xl < x2 < n.In the case of the wave equation the solution u ( x , t ) would remain identically 0 near x = 0 until the "wave" starting at x = xl could move across. For the heat equation the solution u ( x , t ) takes on nonzero values in every x interval for every t > 0. To establish this, we consider a fixed positive t. Then u ( x , t ) can be considered as an analytic function of a complex variable x . By Theorem 26 of Section 9.13, if u ( x , t ) E 0 when x ranges over a whole interval of real values, then u = 0 for this value of t. Hence necessarily, b, = 0 for n = 1 , 2 , . . . , so that f ( x ) = 0 , contrary to assumption. Accordingly, for each positive t , u ( x , t ) takes on nonzero values in every interval of x . The disturbance is propagated instantaneously.

10.11 EQUILIBRIUM AND APPROACH TO EQUILIBRIUM By analogy with case (d) of Sections 10.2 and 10.3 we allow for applied forces F ( x ) and "wall displacements" uo and u , , which do not vary with time. The differential equation and boundary conditions are as follows:

We seek a solution u * ( x )independent of time; u*(x) then describes the equilibrium state of the system. Accordingly, we replace derivatives with respect to t by 0 in (10.103) and are led to the problem

Equation (10.105) is an ordinary differential equation whose general solution is obtained by integrating twice:

If a particular choice G ( x ) of the indefinite integral is made, so that G U ( x ) = -F ( x ) / K * , then the general solution is The boundary conditions give the equations

B+G(O)=uo,

AL+B+G(L)=ul.

Chapter 10 Partial Differential Equations These are easily solved for A and B. The equilibrium state is then

We have assumed F ( x ) to be continuous for 0 5 x 5 L, so that the integrals above have meaning. To describe the approach to equilibrium, we let u(x, t) be an arbitrary solution of (10.103) and (10.104) f o r t > 0 , 0 < I < L. Then y(x, t) = u(x, t ) - u*(x) satisfies the homogeneor~sproblem:

This we verify by substitution in these equations and use of the fact that u(x, t) satisfies (10.103) and (1 0.104), while u*(x) is the solution of the equilibrium problem (10.105) and (10.106). Accordingly,

the general solution is formed of "complementary function" and particular solution in the usual manner. If, in particular, H(x) = 0 and p(x) is a positive constant p , then the complementary function y(x, r ) is a solution of the wave equation; the solutions u(x, t) consist of oscillations about the equilibrium position. If H is also a positive constant, the oscillations are damped (cf. Problem 8 below). If H is so large that p can be neglected, the function y(x, t ) is a solution of the heat equation; equilibrium is approached exponentially.

PROBLEMS 1. Determine the solution for t 2 0,O 5 x 5 n of the heat equation (10.91), with c = 1, such that u(0. t ) = 0, u(n, t) = 0, u(.r, 0) = sinx 5 sin3x. Plot the solution as a

+

function of x and t and compare the rates of decay of the terms in sin x and sin 3x. 2. Determine the solution for t > 0,'O < x < n of the heat equation (10.91) that is continuous for t 2 0.0 5 x 5 IT and has a continuous derivative aulax in this region and that furthermore satisfies the conditions: au/ax = 0 for x = 0 and x = n, u(.r, 0) = f (x), where f ( x ) has continuous first and second derivatives for 0 5 .x 5 n.This can be interpreted as a problem in heat conduction in a slab whose faces are insulated. 3. Determine the solution, for t > 0.0 < x i n,of the equation

such that u ( 0 , t),i&,

u(a,t ) = n:, U(X-, 0) = x

2 sin&.

,

, . l f

689

690

Advanced Calculus, Fifth Edition 4. Prove that if the constants b, are bounded, then the series (10.99)can be written for each t > 0 as a power series in x , converging for all x . [Hint: Let t > 0 be fixed and let 00

bn sin nx cosh ny e-"".".

V ( Xy) , = n=l

Show that the series for v converges uniformly for plying the M-test with

-00

< x < oo,-yl 5 y 5 yl by ap-

I Mn -M~-?"-C-~ >

5.

6.

?

for n sufficiently large. Show that the series remains uniformly convergent after differentiation any number of times with respect to x and y. Each term of the series is harmonic in x and y; hence conclude that v ( x , y ) is harmonic for all x and y . By Section 9.1 1, v ( x , y) can be expanded as a power series in x and y; put y = 0 to obtain the desired series for U ( Xt1.1 , Prove that if the constants bn are bounded, then the series (10.99) remains uniformly convergent for t 2 tl > 0 , -w < x < co,after differentiation any number of times with respect to x and t. r r 1' rl , Prove that if the constants b, are bounded, Ib, I < M, then the function u ( x , t ) defined by (10.99) converges uniformly to 0 as t + oo; that is, given 6 > 0, a to can be found such that lu(x,t)l < E for t > to and -w < x < w. [Hint: Show that lu(x,t)l is less than the (e-rZ')".I sum of the geometric series M Let U ( X , t ) have continuous derivatives through the second order in x and t for t >_ 0 and 0 5 x 5 n and let u(0, t ) = u ( n , t ) = 0. Prove that if u ( x , t ) satisfies the heat equation (10.91) for t > 0 , 0 < x < rr, then u ( x , t ) has the form (10.99). [Hint: See Problem 6 following Section 10.8.1 Discuss the nature of the solutions for 0 < x in , t > 0 of the equation

~ z ,

7.

8.

with boundary conditions u(0, t ) = u ( n ,t ) = 0 , if p , H , and K are positive constants. 9. Prove that if A 0, the equations (10.94) and (10.95) have no solution other than the trivial one: A ( x ) = 0.

We now consider the general problem of type (f): a2u

Ax)at2

a2u + H ( x ) au - - K2= F(x,t), at ax2

0 < x < L, t > 0,

(10.111)

This is the problem of response of the 1-dimensional system to outside forces varying both in position and time. The fact that the boundary conditions (10.1 12) are variable shows that the motion is being forced at the ends x = 0 , x = L also.

Chapter 10 Partial Differential Equations Just as in the case of the equilibrium problem of Section 10.11, we reason that if u(x, t) is a particular solution of (10.1 11) and (10.1 12) for 0 < x < L, t > 0, then the general solution u(x, t) in this domain is

where y ( x , t) is a solution of the homogeneous problem (10.109), (10.1 10). Accordingly, the problem is that of determining a particular solution. We can further concentrate attention on the case of fixed "walls," that is, a(t) = b(t) = 0. To show this, we let

g(x, t) is simply the linear function of x that interpolates between the values a(t) at x = 0, b(t) atx = L. Now (if we assume a(t) and b(t) to have the requisite derivatives) a2g p(x)-

at

+ H(x)-ag - K ~a2g - = G(x, t), at ax2

Accordingly, if u*(x, t) satisfies (10.1 11) and (10.1 12), then w(x, t) = u*(x, t) g(x, t) satisfies ,,

+

Cofiversely, if w(x, t) satisfies (10.114) and (10.1 15), then we verify that w(x, t) g(x, t) = u*(x, t) satisfies (10.1 11) and (10.1 12). Therefore we have reduced our problem to the case of fixed walls. The determination of a function w(x, t) is now accomplished by variation of parameters. The applicability of this method is not dependent on the fact that the coefficients p , H are constant. However, we here confine attention to the case of constant coefficients; the procedure in the general case differs only slightly. Let us suppose first that p = 0 and H = const > 0. Then the "complementary function" is the general solution of the heat equation. We change scale and introduce the abbreviations of Section 10.9, so that the solution is

We now replace the constants bn by functions v,(t) and seek a solution

691

692

Advanced Calculus, Fifth Edition

2

We proceed formally and then determine conditions under which the solution obtained is valid. Substitution of (10.1 16) in (10.1 14) (with p = 0 ) leads to the equation CIO ,tS: i

H n=l

{v~(t)e-""" sinnx} = ~ ~ r( ) ;x , B

:i'F

r t

the other terms cancel out. This equation is simply a Fourier sine series in x for Fl ( x , t ) . Hence a I,. rs: H d".e-n?clt - Fl(x. t ) sinnx dx. dr 0 (10.117) vn( r ) = en''".( Fl ( x ,t ) sin nx dx nH

/"

1

1%

/

Since we seek only one particular solution, we can choose the indefinite integral here so that vn(0)= 0 ; that is, we choose

2

1' { ln

vn(t)= nH

8 ' ~ ' ~ F, (x, s ) sin nx d x

where s is a dummy variable of integration. Accordingly, vn(t)=

1'/"

1 nH

0

.!I

~ ( xs ),en''" sin nx dx ds,

(10.118)

0

and the particular solution sought is W ( X , I)=

2

-

O0

n H 11=I

( sin

nx e-"""

/' 0

Fl(r,s) snliSsinnx dx ds 0

We now study the validity of the result. If F l ( x ,t ) is defined for t L 0 and 0 (x 5 n and is continuous in both variables in this region, then v,,(t)is well defined by (10.1 18). Furthermore, let a'FI/ax2 be continuous in x and t , so that a2Fl/ax2 has a maximum M ( t l )in each rectangle: 0 ( x 5 n , 0 ( t I t l ;let also FI(O,t ) = F l ( x ,t ) = 0. Then, as in Section 7.8, we conclude that for 0 ( t ( tl

Hence by (10.118), in this rectangle,

,

f

i?gi:

and each term of the series (10.1 19) is bounded by 2 ~ ( t I ) / c ~The n ~series ~ . therefore converges uniformly and continues to do so after differentiation once with respect to t or twice with respect to x . We therefore find H-aw - K -,a2w =at ax2 n

2 l" sin nx

F1(r, r) sin nr dr.

Chapter 10 Partial Differential Equations The right-hand side is the Fourier sine series of Fl( x , t) and, under the assumptions made, converges to Fl ( x , t). Hence (10.114) is satisfied, as is (10.1 15). Therefore (10.119)is indeed the particular solution sought. The function

is then a particular solution of ( l o .1 11) and (10.1 12), and

is the general solution, where y(x, t ) =

i",P b, sin nx e-n2c2t,

the b, being arbitrary constants. To determine a solution satisfying the initial condition u(x, 0)= f ( x ) ,we have to determine the constants b, so that

that is, the series

b, sin nx must be the Fourier sine series of

We have throughout assumed p = 0 and H = const. If p is a positive constant and H = 0,then the complementary function is the general solution of the wave equation:

2

sin nx(u, sin nnt

+ j3, cos nat).

n= l

The trial function (10.1 16) is now replaced by the function:

Substitution in (10.1 14) does not yield enough conditions to determine p,(t) and q,(t), because the equation is now of second order in t. We choose an extra set of conditions:

These conditions can be obtained by replacing (10.114) by a system of equations:

693

694

Advanced Calculus, Fifth Edition (Problem 5). Proceeding as before, we find

1'lH

2 pn(t) = nun/'

0

5 naJrP

0

qn(t) =

F,(X.s ) cos n a s sinnx d x d r ,

Jt

(10.122)

o

F ~ ( Xs, ) s i n n a s sin nx d x ds.

A similar procedure applies when p and H are both positive constants.

PROBLEMS 1. Find the solution of the partial differential equation

satisfying boundary conditions u(0, t) = 0, u(n, t) = n 2 sint, and initial conditions u(x, 0) = n x - x2. 2. Let U(X,t) be a solution of the partial differential equation

and boundary conditions u(0, t) = 0, u(n, t) = 0, u(x, 0) = 0, au/at(x, 0) = 0. Show that resonance occurs only when w = f1 and determine the form of the solution in the two cases: w = f 1, w # f1. [The other resonant frequencies 2, 3, . . . are not excited because the force F(x, t) is orthogonal to the corresponding "basis vectors" sin 2x, sin 3x, . . . ;cf. Problem 9 following Section 10.3.1 3. Let the outside force F(x, t) be given as a Fourier sine series:

Obtain a particular solution of the partial differential equation

with boundary conditions u(0, t ) = 0, u(n, t) = 0, by setting n=l substituting in the differential equation, and comparing coefficients of sin nx (corollary to Theorem 1, Section 7.2). Show that the result obtained agrecs with (10.1 19). 4. Show that the substitution of (10.116') in (10.1 14), with p > 0, H = 0, and application of (10.120) leads to the Eqs. (10.122) for pn(t),qn(t). 5. Show that the solution of the homogeneous problem (F = 0) corresponding to (10.121), with boundary conditions w(0, t) = 0, w(n, t) = 0, is given by 00

w = E s i n n x [ a n sinnat

+ bn cosnat],

n=l 00

z

sin nx[nacrn cos nut - nab,, sin nat].

=

n= 1

Chapter 10 Partial Differential Equations Show that replacement of a, by p,(t), p, by q , ( t ) , and substitution in (10.121) lead to the equations: sinnat + q; cos nut = 0, nap; cos nut - naq; sin nut = nP and hence one obtains (10.122).

S"

Fl sin nx d x ,

0

6. Let u 1 ( x , t ) , u 2 ( x ,t ) , u 3 ( x ,t ) , respectively, be solutions of the problems (for 0 < x < n, t > 0): 4

.

U,

Show that u l ( x , t )

- UIX = F(x,t),

u(n,t ) = 0 ;

u(0,t ) = 0 ,

+ u2(x, t ) + u 3 ( x ,t ) is a solution of the problem

u, - u,, = F ( x , t ) ,

u(0,t ) = a ( t ) ,

'

u ( n , t ) = b(t).

This shows that the effects of the different ways of forcing the system combine by superposition.

10.13 EQUATIONS WITH VARIABLE COEFFICIENTS 8 STURM-LIOUVILLE PROBLEMS In order to determine the normal modes in the problem

we make the substitution: u = A ( x ) sin(ht

+E )

and are led to the equations:

When p ( x ) is constant, we know that the only solutions of (10.124) are the functions A,,(x) = c sin ( n n x l L ) ;the associated frequencies A, are of form a n . What is the nature of the solutions when p is variable? A similar question arises if we consider the problem for the heat equation with variable coefficient H ( x ) :

695

696

Advanced Calculus, Fifth Edition The substitution u = ~(x)e''

leads to the equations

Except for a change in notation, these equations are the same as (10.124). The problems (10.124) and (10.126) are special cases of the class of SrurmLiouville boundary value problems. A more general case is as follows:

Here the function y(x) is to be a solution of the differential equation for a 5 x 5 b and is to satisfy the given boundary conditions at a and b. We further assume that r(x), p(x), q(x) have continuous derivatives over the interval and that r(x) > 0, p(x) > 0. A value of h for which (10.127) has a solution other than y(x) r 0 is called a characteristic value. One could allow for complex characteristic values, but it can be shown that under the assumptions made this does not arise; hence we restrict to real-characteristic values. For each characteristic value A there is an associated solution y ( x ) , called a characteristic function; the functions cy(x), where c is a constant, are also characteristic functions. It could conceivably happen that there are functions besides cy(x) that have the same h; it can be shown that under the assumptions made this cannot arise: THEOREM The characteristic values of the Sturm-Liouville problem (10.127) can be numbered to form an increasing sequence: A, c h2 < . . . < A, < . . . .The corresponding characteristic functions can be numbered similarly to form a sequence: yn(x); each y,(x) is determined only up to a constant multiplier. The functions y,(?c) are orthogonal with respect to the weight function p(x):

The Fourier series of a function F(x) with respect to the orthogonal system { m y , ( x ) } converges uniformly to F(x) for every function F(x) having a continuous derivative for a ( x 5 b and such that F ( a ) = 0, F(b) = 0. '

For a proof of this theorem and of theorems for more general Sturm-Liouville boundary value problems, we refer to the books of Hellwig, Titchmarsh (1962), and Kamke listed at the end of this chapter. (See also Problem 13 following Section 2.18.) Because of the theorem, we can be assured that except for minor changes in form, the statements about the wave equation and heat equation in Sections 10.7 and 10.9 continue to hold when the coefficients p(x), H(x) are variable [p(x) > 0,

Chapter 10 Partial Differential Equations H ( x ) > 01. For example, the characteristic functions A n ( x ) of (10.124) pravide normal modes: .

A n ( x )sin (Ant

+6 )

and constant multiples thereof. The "general solution" of (10.123) is again a series m 00 7 A . ( r ) [ a , sin Ant 8. cos Ant]. c n A , ( x )sin ( l n r 6.) = n=l n=l To satisfy initial conditions:

+

+

one has only to choose the constants a,,, pn so that

this requires expansion of the functions

f ( x ) , m g ( x ) in Fourier series:

By the preceding theorem these expansions have the same properties as the sine series used above, so that there is no change in the results. The theory of equilibrium states, approach to equilibrium, and forced motion can also be repeated. The only difficulty is in effective determination of characteristic values and functions. For various special equations, infinite series are effective. For others, one is forced to use numerical methods. These are discussed in Sections 10.16 and 10.17. The previous theorem on the Sturm-Liouville problem can be extended under appropriate assumptions to the "singular case" in which the function r ( x ) is 0 at a or b or both, while remaining positive for a < x < b. This case includes in particular the important problem:

+

whose solutions are the Legendre polynomials Pn(x),with An = n(n 1) (Section 7.14.No boundary condition is imposed at x = f1 , but it is required that the solutions remain continuous at these points. The theory can also be extended to include the problem:

The solution is required to be continuous at x = 0 . The solutions of (10.129) are the where s, ranges over the positive roots of the Bessel function functions Jm(smnx), J,(x); the corresponding Am, = s;,. The extensions to these cases are covered in the book of Titchmarsh (1 9 6 2 ) listed at the end of the chapter. . -

697

698

Advanced Calculus, Fifth Edition

10.14 EQUATIONS IN TWOAND THREE DIMENSIONS 8 SEPARATION OF VARIABLES The generalization of our basic problem to two and three dimensions brings no basic change in the results, though the determination of normal modes is in general more complicated. As an exampje we consider the wave equation for a rectangle:

The substitution

leads to the characteristic value problem

A(x, y) = 0 forx = 0, x = n, y = 0, y = n.

In order to determine the characteristic functions A(x, y) we seek particular characteristic functions having the form of a product of a function of x by a function of y:

Accordingly,

X"

y"

-+-+,=o. X Y

~2

a Now if we vary x, the second and third terms of the last equation cannot vary; hence the first term is a constant. Similarly, the second term is a constant:

Because of the boundary conditions in (10.131), we are led to two new boundary value problems:

The first is satisfied by X,(x) = sinnx, for p = n2; for this value of p the second is satisfied by the functions Y,(y) = sinmy, provided that (h2/a2) - p = m 2 .

Chapter 10 Partial Differential Equations Accordingly, Amn(x,y) = sin nx sin my

+

= a2(m2 n2), where m and'n are integers. For a is a solution of (10.131), for given characteristic value h there may be several characteristic functions; in fact, if A,,(x, y) is one function, then Anm(x,y) provides a second one, unless m = n. This causes no trouble because we put all linear combinations into the "general solution":

+ c).

cmnsinnx sin my sin(Amnt

U(X,y , f) =

The series is a "double series," to be summed over all combinations of positive integral values of rn and n ; the series is a Fourier series in the two variables x , y and, as was pointed out in Section 7.16, can be rearranged to form a single series or summed as an "iterated" series:

2[5

+ 6.1

cmnsin nx sin my sin (hmnt

m=l

n=l

Since only sine functions appear, the series is really a Fourier sine series in two variables. As in Section 7.16, the functions sinnx sinmy form a complete orthogonal system for the region: 0 5 x (n, 0 5 y 5 n.Accordingly, the results achieved in one dimension can all be generalized to two dimensions. Because of the more complicated characteristic values, the solutions are more difficult to analyze; in particular, the solutions are in general no longer periodic in t. The crucial step was the replacement of A(x, y) by the product X(x)Y(y). This led to a so-called separation of variables in (10.133) and determination of particular characteristic functions A(x, y), which together form a complete orthogonal system. The fact that such a procedure can be successful was already indicated in our method for determining normal modes for the problems in one dimension; the substitution u = A(x)e" could have been replaced by a substitution u = A(x)T(t), and a separation of variables would then have led to the same results. The method of separation of variables thus appears as a general method for attacking homogeneous linear partial differential equations. The method may in some cases provide only certain particular solutions; in a wide variety of cases these particular solutions have been shown to provide a complete set of orthogonal functions. These cases include the Laplace equation in cylindrical and spherical coordinates (Sections 2.17 and 3.8); see Problems 4 and 7 following Section 10.15. The equilibrium problem: V2u = -F ( x , y), with values of u prescribed on the boundary of a region R of the xy-plane, can be attacked in several ways. It can be shown that under appropriate assumptions on F(x, y ) the function (logarithmic potential) F(r, S)log [(x - r)2

+ (y - s ) ~d]r ds

(10.135)

satisfies the Poisson equation V2uo= - F inside R and is continuous in R plus boundary. The function v = u -uo will then satisfy the condition v2v= - F F = 0

+

699

700

Advanced Calculus, Fifth Edition

-

inside R and have certain new boundary values on the boundary of R. Determination of v is then a Dirichlet problem, which can be attacked by conformal mapping (Section 8.22). While conformal mapping is not available as a tool for the corresponding problems in three dimensions, methods based on potential theory can be used; see especially the book of Kellogg listed at the end of the chapter. (An introduction to potential theory is given in Sections 5.16 to 5.18.) One can in general reduce the equilibrium problem v 2 u = -F to the case of zero boundary values by the following procedure (cf. Section 10.12): Let u(x, y) be required to have values h(x, y) on the boundary C of R. If h(x, y ) is sufficiently smooth, one can then find a function h (x, y) that has continuous first and second derivatives inside R, is continuous in R plus C, and equals h(x, y) on C. The function v = u - h l is then zero on C, and V2v = - F - V2hl = - F1(x, y). The determination of v can be carried out with the aid of a Green S function, as will be indicated in Section 10.18. (See also Sections 5.17 and 5.18.)

In many physical problems it is natural to regard the continuous medium as being injnite in extent. For example, in one dimension, one can consider the wave equation

for the infinite interval x > 0. If one seeks normal modes, one is led to the characteristic value problem:

This problem has solutions for every value of A, namely, the functions sinax, for a 2 a 2 = h2. Thus the resonant frequencies A form a "continuous" set of numbers, and one has a "continuous spectrum." [There also exist unbounded "normal modes": u = sinh a.3: eUa'.These are of less physical interest.] One can construct linear combinations of the normal modes in order to obtain a "general solution" of the homogeneous problem. Since there is a continuous sequence of A's, an integration is called for rather than a summation. For a 1 0 we must integrate expressions of form sin x[p(a) cos ( a a t )

+ q ( a ) sin (aat)],

where p ( a ) and q(a) are "arbitrary" functions of a . We obtain the integral

lrn

sinux[p(a) cas (aar)

+ q(a) sin (oar)] da.

For each fixed t this can be considered as a Fourier integral (the Fourier sine integral, Section 7.18). In particular, for t = 0 we obtain a Fourier integral representation of the initial displacement u(x, 0):

lrn

p ( a ) sin a x do.

-

..

Chapter 10 Partial Differential Equations Since the theory of the Fourier integral has been highly developed, most of the results for the finite interval can be extended to the infinite case. Similar statements can be made for problems in two or three dimensions for unbounded regions. For more information w e refer to the books of the following authors, as listed at the end of the chapter: Sneddon, Titchmarsh (1948), Wiener, Courant and Hilbert, Frank and von Mises, and Tamarkin and Feller. T h e Laplace transform (Section 6.25) can also be used to represent solutions over infinite intervals. This is discussed in the texts mentioned and in the second book by Churchill (1972).

PROBLEMS 1. a) Let a vibrating string be stretched between x = 0 and x = 1; let the tension IC2 be (x I ) and ~ the density p be 1 in appropriate units. Show that the normal modes are given by the functions

+

I

n2Tr2

All(x) = J Z - l s i n

log 2

[Hint: Make the substitution x + 1 = eUin the boundary value problem for A,,(x).] b) Show directly that every function f (x) having continuous first and second derivatives for 0 5 x 5 1 and such that f (0) = f ( I ) = 0 can be expanded in a uniformly convergent series in the characteristic functions A,(x) of part (a). [Hlnt: Let x 1 = eU as in part (a). Then expand F(u) = f (eu - 1)e- f ' in a Fourier sine series for the interval 0 5 u _( log2.1 2. Show that the general second-order linear equation

+

where po(x) # 0, takes on the form of a Sturm-Liouville equation (10.127) if the equation is multiplied by r(x)/pa(x), where r(x) is chosen so that r ' l r = pl/po. In general, an equation of form: (ry')' h(x)v = 0 is called self-adjoint. 3. Obtain the general solution, for t > 0 . 0 < x < n,0 < y < n, of the heat equation with boundary conditions:

+

4. Show that separation of variables u(r, 0) = R(r)@(O)in the problem in polar coordinates for the domain r < 1:

leads to the problems:

If we require that u(r, 8 ) be continuou< in the circle r 5 1, then 0 ( 6 ) must be periodic in 0, with period 2n. Show that this implies that p = m2(m = a, 1.2.3, . . .)-and that

Advanced Calculus, Fifth Edition R(r) = ~ , ( & r ) , functions

for I = I,,;

cf. (10.129). Hence one obtains the characteristic

J , ( ~ ~ ) c ome, s

J,(&r)

sin me,

and linear combinations thereof. It can be shown that these characteristic functions form a complete orthogonal system for the circle r 5 1. 5. Using the results of Problem 4, determine the normal modes for the vibrations of a circular mcmbranc; that is, find normal modes for the equation

6. Using the results of Problem 4, determine the general solution of the heat conduction problem:

U(X,y, t ) = 0

for x2

+ y2 = 1.

7. Show that the substitution u = R(p)@(@)0(8)in the problem in spherical coordinates for the domain p < 1: v 2 u + Au

[

1 ., a =sin @ P2 sin2 @ a~

(p ,a,)a~

u(p, @ , e l= 0

(sin@-- a,)

a a@

+sin@-

a@

+-a2,] ae2

+Iu=O,

for p = 1 ,

leads to the separate Sturm-Liouville problems:

+

( p 2 ~ ' ) ' (Ap2 - a ) R = 0,

R =0

(sinq5@')'+(asin@-,9csc@)@=O,

for p = 1; 0"+,9@=0.

Here a,B, and h are characteristic values to be determined. The condition that u be continuous throughout the sphere requires O to have period 237, so that B = k2 (k = 0, 1,2, . . .) and Ok(0) is a linear combination of cos kB and sin kB. When B = k2, it can be shown that continuous solutions of the second problem for 0 5 @ 5 rr are obtainabfe only when a = n(n l), k = 0, 1, . . . , n, and 4, is a constant times Pn,k(cos@),where

+

+

and Pn(x) is the nth Legendre polynomial. When a = n(n 1) (n = 0, 1, 2, . . .), the first problem has a solution continuous for p = 0 only when h is one of the roots &+f.17An+:,2?. . . of the function J,+: (fi), where J,+; (x) is the Bessel function of order n f ; for each such I the solution is a constant times p-: J ~ + ; ( A ~ Thus ) . one obtains the characteristic functions p-i Jn+;(fip)~,,k(cos@) cos k ~p-i , /,+;(fi p)pn,l (COS@) sin k0, where n = 0, 1 , 2 , . . . , k = 0, 1, . . . , n, and h is chosen as above for each n. It can be shown that these functions form a complete orthogonal system for the spherical region p 5 1. 8. Show that the wave equation (10.136) for the infinite interval -00 < x < oo has a continuous spectrum and find the characteristic functions for the bounded normal modes.

+

Chapter 10 Partial Differential Equations

For problems with variable coefficients or problems in two or three dimensions concerning regions of inappropriate shape, the methods described before will in general fail to produce solutions in a form suitable for numerical applications. Similar remarks apply to classes of differential equations more general than those considered here, in particular nonlinear equations. Although the theoretical aspects of the subject are highly developed and one can often establish existence of solutions, this is not always sufficient for the needs of physics. (Such an existence theorem is the famous Cauchy-Kowaleski theorem for initial value problems; see Chapter 3 of the second book by John listed at the end of the chapter.) Accordingly, a variety of numerical methods have been devised for explicit determination of solutions satisfying given boundary conditions and initial conditions. We consider briefly some of these methods. The first method consists simply in a reversal of the limitprocess of Section 10.5. We replace the derivative a2u/ax2by the difference expression

where u, = u(x,). From the differential equation

we are thus led to the system of equations d2u, du, m,h, - - k dt2 dt where a = 1 , . . . , N and

+

(&+I

-

.

- 2~4, + u,-~) = F,(t),

(10.139)

Equations (10.139) can be handled completely by the methods of Section 9.6. The tools required are basically algebraic, in particular the solution of simultaneous equations. In order that (10.139) be an accurate approximation to (10.138) it is necessary that N be large; this makes the algebraic problems far from trivial, at least as far as time requirements are concerned. Initial value problems. If one seeks a particular solution of (10.138) satisfying ) ,can given initial conditions and boundary conditions (values of uo and u ~ + ~one set up the approximating equations (10.139) and solve these numerically (see the book by Isaacson and Keller and that by Iserles listed at the end of the chapter). These measures are even more appropriate if nonlinearity is present, for example, if a2u/ax2 is replaced by its square. Characteristic value problems. Determination of normal modes for a wave equation or heat equation obtained from (10.138) leads in general to a Sturm-Liouville problem (10.127). This can be attacked by considering the approximating problem (10.139), for which determination of normal modes is an algebraic problem; one

703

704

Advanced Calculus, Fifth Edition can also use difference equations, as in Problems 5 to 10 following Section 10.4. A variational method is also helpful; this is described in Section 10.17. One can treat the problem as an initial value problem, in the following way. To solve the equations: one constructs particular solutions of the initial value problem: A(0) = 0,A'(0) = 1 for different values of A. As h is gradually increased, the solutions vary in a simple manner, and one can by trial and error determine solutions for which the condition A(l) = 0 is satisfied. These are precisely the characteristic functions sought. Equilibrium problems. The equilibrium problem for (10.138) is solved in all generality in Section 10.1 1; the only difficult step is an integration, which might have to be carried out numerically as in Section 4.2. Another way of writing the solutions, with the aid of a Green's function, is explained in Section 10.18. Problems in two dimensions. If a2u/ax' is replaced by a 2-dimensional Laplacian v2uin (10.138). so that one has a problem for u(x, y , t ) in a region R of the xy-plane, an approximating system analogous to (1 0.139) can be devised. If R is a rectangle a ( x 5 b, c 5 y 5 d, one can divide R into squares (if the sides of R are commensurable) of side h. We then consider the values of u only at the corners of the squares. At each such comer (x, y) the Laplacian is computed approximately (Problem 5 following Section 6.2 1) as the expression u(x

+ h, y ) + u(x, y + h) +

U ( X - h,

y)

+ U(X,y - h) - 4u(x, y)

h' A system of equations analogous to (10.138) is obtained. If R is not a rectangle, one can approximate R by a figure pieced together of rectangles and proceed similarly. The statements concerning the initial value problem above can now be repeated without change. The characteristic value problem can also be replaced by an approximating algebraic problem in the same way; the variational approach of Section 10.17 is also useful. The equilibrium problem can be attacked numerically by considering the approximating system of equations in variables u, as above. One has then N simultaneous linear equations; if N is large, these may be far from simple to handle. One can also regard the equilibrium problem as a special case of a heat equation: u, - K'V'U = F(x,y), with u given on the boundary of R, for all solutions of the heat equation tend exponentially to the equilibrium solution. One can assign arbitrary initial values and obtain a particular solution; for t large this will approximate the equilibrium solution sought. The variational methods of Section 10.17 are also of use for the equilibrium problem. Most of the remarks made can be generalized to problems in three diinensions. The accuracy of the approximate methods described has been investigated, and in general the processes described can be carried out to yield the accuracy desired for the solutions. Some details of this are given in the books of Tamarkin and Feller, Isaacson and Keller, Iserles, and John (1967) listed at the end of the chapter. It is worth remarking that the crucial step of replacing (10.138) by (10.139) can be regarded as the replacement of one physical model bv another. According to modern

705

Chapter 10 Partial Differential Equations physics, both models are an oversimplification of what is observed in nature. If either one serves to describe the phenomena concerned with sufficient accuracy, then it can be regarded as a useful one.

The equilibrium solution for systems (10.139) and the analogous ones in two and three dimensions can be considered as problems of minimizing a function $ ( u l , . . . ,u N ) .For, as was remarked in Section 10.4, Eqs. (10.139) can be written in the form d 2 ~ , dun av ma-+ha-+-= F,(t) ( a = 1 ,..., N). (10.141) dt2 dt au, The equilibrium problem is then the problem

2

rl

where the F, are constants. The end-values uo and U N + I are also given as constants; we can consider them to be 0 by modifying the definition of Fl and FN.If we now let

+ - - .+ F N U N ) ,

~ ( u I ., .., U N ) = V ( U I ., .. , U N ) - (Flu1

then (10.142)is simply the condition that

-

(10.143)

~ l c j ,

For (10.139) we have

.

and we can verify that (10.144) has precisely one solution u;, . ., u*, and that this critical point is a minimum point is also easily verified (Problem 6 following Section 10.18).A similar statement applies to the analogous problems in two or three dimensions. Indeed, the physical picture that leads to equations of form (10.141)is almost always that of a system of particles capable of an equilibrium state, at which the potential energy V has its smallest value; Eqs. (10.141)then describe the forced oscillations about this equilibrium state. When the applied forces are constant, the potential energy V is replaced by a modified one 4; the Eqs. (10.141) then become

the new equilibrium state is now the minimum of $. By appropriate passage to the limit it can be shown that the equilibrium problem for (10.138) is equivalent to minimizing the expression

706

Advanced Calculus, Fifth Edition The expression @, is afunctional, that is, an expression whose value depends on the function u(x) chosen. It can then be shown that the function u*(x) that satisfies the equilibrium conditions:

assigns to @, its smallest value attainable among all smooth functions u(x) satisfying the boundary conditions (Problem 7 following Section 10.18). The general problem of minimizing functionals such as @, is the subject of the calculus of variations. Accordingly, methods based on minimizing (or maximizing) appropriate functionals or functions are called variational methods. One can attack the problem of minimizing the functional @, of (10.147) by the following procedure, due to Rayleigh and Ritz. One chooses a particular function U(X)depending linearly on several arbitrary constants:

The functions ul(x), . . . , u,(x) are chosen to satisfy the boundary conditions, that is, to be 0 at x = 0 and x = L; they are otherwise chosen as desired, though the effectiveness of the method depends greatly on the skill with which they are chosen. On substituting the expression (10.149) in (10.147), one obtains a function P whose value depends only on the constants cl, . . . , c,. Because of the form of @,, P(cl, . . . , c,) is also a quadratic expression, and its minimum is the unique solution of the equations

These are simultaneous linear equations. On solving for cl , c2, . . . , c,, one has found a function (10.149) that gives @, a smaller value than that given by certain competing functions. If the class of competing functions is large enough, one can expect the function u(x) found to be close to the true minimum of a,. The functions u j(x) can be chosen as piecewise linear or piecewise polynomial functions. This is the basis of the widely usedjnite element method. For the equilibrium problem in one dimension the procedure described is not needed, for one can solve (10.105) explicitly as in Section 10.11. However, for problems in two and three dimensions, explicit solution is usually difficult (cf. Section 10.14), and the Rayleigh-Ritz procedure can be of considerable value. It can be shown that the solution of the equilibrium problem -K * V ~ U= F in two and three dimensions is equivalent to minimizing the functionals

respectively.

Chapter 10 Partial Differential Equations

707

Variational methods can also be applied to the determination of characteristic values and functions. For example, the characteristic value problem for the normal modes of (10.139),with h, = 0 and F, = 0, is the problem h 2 m , ~ , + k 2 ( ~ , + 1 - 2 A , + A , - I ) = 0 ( u = l , ..., N),

(10.152)

where A. = A N + ] = 0. These are the equations for minimizing the function V(AI,..., A N ) = k 2 ( ~ : + . - . + A ~ - A ~ A ~ - .A.N- --I A N )

(10.153)

subject to the side condition 1 (10.154) i 1) = 0. g(A1,.. . , A N ) = - ( m , ~ : + + m N ~ 2 Indeed, the method of Lagrange multipliers for this problem (Section 2.20) gives the equations

which are the same as (10.152). The side condition (10.154) fixes the constant of proportionality of the A's (up to a f sign). The characteristic values A,, . . . , AN correspond to critical points of V on the "ellipsoid defined by (10.154).In general, two of the A's correspond to the absolute minimum and maximum of V, when (10.154)holds. Again a limit process leads to a variational formulation of the characteristic value problem for the continuous medium. For the equation

the characteristic value problem concerns the "critical points" of the functional

subject to the side condition -

R

-

The Rayleigh-Ritz method is applicable here in the same way as above. For further information on this topic, one is referred to the article and books by Courant, Ciarlet, Gould, Iserles, and Kantorovich and Krylov listed at the end of the chapter.

Given the equilibrium problem

+

-kZ(~,+I- 2 ~ , , u , - ~ ) = F,

(U = 1, . . . , )

Uo

= U N + ~= 0, (10.155)

one could obtain the solution by the following method. One could first solve the problem for which Fl = 1 and Fz = F3 = F4 = . . = 0; let the solution be u,, = g,,, 1.

-

.

708

Advanced Calculus, Fifth Edition One could then solve for F2= 1 and all other F, = 0, obtaining go,2; in general u, = g,,, is the solution when F, = 1 and Fa = 0 for a # p. The linear combination

is then the desired solution of (10.155); for substitution of u, in the left-hand side of the first equation (10.155) gives F , , since g , l gives 1 and all other g's give 0; a similar reasoning holds for the other equations. Thus the effect of all the Fa can be built up by superposition of unit forces. By a limit process a similar result is obtained for the problem:

One finds (10.158) where the g(x, s) are the solutions for a force F "concentrated at a point s." The function g(x, s ) is called the GreenSfinction for (10.157); it is 0 when x = 0 and x = L, has the value s(L - s)/L when x = s, and is linear in x between these values. Accordingly, g(x, s ) has a "corner" at x = s , due to the concentrated force at this point, whereas aZg/ax2= 0 otherwise. Similar results hold for quite general nonhomogeneous linear equations. In particular, a Green's function g(x, y; r, s) can be found for the problem (Poisson equation): t

.

-

v 2 u = -F(x, y) inside R ,

(10.159)

p

u(x, y) = 0 onboundaryof R, for a general region R in the plane. The solutions of (10.159) are then given by a formula: U(X,y) =

11

(10.160)

g ( x , y;r, s)F(r. s) d r ds.

R

For each (r, s ) the function g satisfies the equation v 2 g = 0 except at x = s, y = r , where it has a discontinuity corresponding to a "point load." Also g(x, y; r, s) = 0 when (x, y) is on the boundary of R; therefore u is given as a "linear combination" of functions all 0 on the boundary of R and is therefore itself 0 on the boundary. (See Section 5.17.) In order to solve the characteristic value problem v2u +Au = 0 in R,

'

'

, -

8

r-.

(10.161) u = 0 on the boundary of R, one can rewrite the problem in form (10.159): V2u = -hu; therefore ufx, y) = A

SJ

g(x, y; r, s)u(r, s ) d r ds.

Chapter 10 Partial Differential Equations This gives an implicit equation for u, the crucial operation being integration. The equation is called an integral equation. A variety of other problems in partial differential equations can be restated as integral equations. A number of methods are available for obtaining solutions of integral equations, and they must be considered among the most powerful ways of attacking partial differential equations. The following features are of great importance: The theory of integral equations is much more un@ed than that of differential equations, problems in one, two, or three dimensions being treated in the same way; the treatment of boundary values is simpler, for example, in (10.162) the boundary condition on u is automatically taken care of, since g = 0 on the boundary; the methods of solution are much more adaptable to nonlinear problems. For a full discussion of integral equations and their applications refer to the books of Tamarkin and Feller, Frank and von Mises, Courant, Kellogg, and Mikhlin listed at the end of the chapter.

-

<

PROBLEMS 1. Let the equilibriumproblem v2u(x, y) = 0 be given for the square 0 Ix 5 3 , 0 Iy 5 3, withboundaryvaluesu=x2fory = 0 , u = x 2 - 9 f o r y = 3 , ~ = - ~ ~ f o r x = O , u = 9 - y2 for x = 3. Obtain the solution by considering the heat equation u, - v2u = 0.Use only integer valqes of x, y so that only four points (1, l), (2, I), (1, 2), (2, 2) inside the rectangle are concerned. Let ul, u2, u3, u4, respectively, be the four values of u at these points. Using the given boundary values, show that the approximating equations are

+

.

Replace by difference equations in t: Au 1 = (u2 us - 4ul)At, . . , where Aui = ui(r At) - ui(t). These equations can be used to obtain ul, . . . , u4 numerically at to At, to+2At, .. .from given initial values at to (Euler method). Take to = 0, At = 0.1, and ui(0) = 1 for i = 1, . . . , 4 to find ui(l). Verify that the values found are close to the equilibrium values: ul = 0,242 = 3, us = -3, u4 = 0. 2. The method of relaxation or Liebmann's method, as applied to Problem 1, consists in choosing initial values of ul, u2, u3, u4, then correcting each one in turn by replacing it by the average of the four neighboring values. Thus at (1, 1) the value ul would be replaced by the average of u2, us, - 1, and 1. The value u2 at (2,l) would then be replaced by the average of 8, u4, ul (new value), and 4. Apply this process repeatedly, starting with ul = u2 = u3 = u4 = 1 and show that the corrected values gradually approach the equilibrium sought. This technique is discussed in the two books of Southwell listed at the end of the chapter. 3. Let the wave equation problem u,, - v2u = 0, U(X,y, t) = 0 on the boundary, be given for the square of Problem 1. Determine the resonant frequencies by using difference expressions for v2u as in Problem 1, so that one has the equations:

+

+

The exact frequencies are found as in Section 10.14 to be in(rn2 + n2)f (m = 1 , 2 , . . . , n = 1,2, . . .). Show that the four lowest frequencies are fairly well approximated. 4. Let (10.138) be the wave equation u,, - a2u,, = O for the interval 0 < x < rr as in Section 10.7. The corresponding approximating system (10.139) was considered in

709

710

Advanced Calculus, Fifth Edition

"1'

Problem 7 following Section 10.4. In the notation used here the characteristic values and functions found were 2a(N 1) sin nn I, = A, (x,) = sin (nx,), n 2 ( N 1) ' whereu = 0 , 1 , ..., N l , n = 1 , . .. , N;comparewiththeexactsolutionsofthewave equation. Show that for each fixed n , A, -+ an as N + CQ. 5. Study the behavior of the solutions of the initial value problem:

+

+

+

as I increases from 0 to CQ; note in particular the appearance of values of 1 for which the condition u ( 1 ) = 0 is satisfied. It can be shown that the same qualitative picture holds for . the general Sturm-Liouville problem of Section 10.13. 6. Show that the function 4 defined by (10.145) has precisely one critical point at which + 4 takes on its absolute minimum. [Hint: Show that by proper choice of the constants, cul, . . . , ( Y N , the substitution

transforms 4 into an expression

+ +

This shows that 4 + CQ as w: . . . W; + 00, SO that 6 has at least one critical point that gives its absolute minimum. The equations in ul , . . . , u ~ for the critical point are simultaneous linear equations. The equations have a unique solution, if the corresponding homogeneous equations (all F, zero) have a unique solution; in the homogeneous case ul = u2 = . . . = U N = 0 is one critical point; if u;, . . . , U ; were a second one, then a @ / a u , would be 0 for all u when u l = u ; t , . . . , u N = u&t and -oo c t c CQ. This contradicts the fact that 4 + oo as w: - .. w: -+ CQ. The uniqueness of the critical point can also be established by using difference equations, as in Problem 5 following Section 10.4.1 7. Prove that the function Q, defined by (10.147) attains its minimum value, among smooth functions u ( x )satisfying the boundary conditions u ( 0 ) = u ( L ) = 0, when u is the solution of the equation -K2u"(x) = F ( x ) . [Hint: Take L = n for convenience. Then express the integral in terms of Fourier sine coefficients of u ( x ) ,u l ( x ) ,and F ( x ) , using Theorem 14 of Section 7.13. This gives a separate minimum problem for each n, which is solved precisely when -K2u" = F ( x ) . ] 8. a) Determine the function u ( x ) that minimizes

+ +

b) Use the Rayleigh-Ritz procedure to solve the problem of part (a), using as trial functions the functions

9. Verify that the Green's function for (10.157), as described in the text, is the following function when L = 1:

Chapter 10 Partial Differential Equations

711

Verify that the function

solves the problem u N ( x )= - x , u(0) = u(1) = 0. Suggested References Bateman, H., Partial DtTerential Equations of Mathematical Physics. New York: Dover, 1944. Churchill, R. V., Operational Mathematics, 3rd ed. New York: McGraw-Hill, 1972. Churchill, R. V., and James W. Brown, Fourier Series and Boundary Value Problems, 5th ed. New York: McGraw-Hill, 1993. Ciarlet, P. G., Numerical Analysis of the Finite Element Method. Montreal: Les Presses de 1'UniversitC de Montrtal, 1976. Courant, R., Advanced Methods in Applied Mathematics. Lithoprinted notes of lectures at New York University, 1941. Courant, R., VariationalMethodsfor the Solution of Problems of Equilibrium and Vibrations. Bulletin of the American Mathematical Society, vol. 49, pp. 1-23, 1943. Courant, R., and D. Hilbert, Methods ofMathematica1Physics, 2 vols. New York: Interscience, 1989. Doetsch, G., Theory and Application of the Laplace Transformation (trans. by W. Nader). Berlin: Springer, 1974. Frank, P., and R. von Mises, Die Dzyerentialgleichungen und Integralgleichungen der Mechanik und Physik, 2 vols., 2nd ed. New York: Dover, 1961. Goldstein, Herbert, Classical Mechanics, 2nd ed. Reading, Mass.: Addison-Wesley, 1980. Gould, S. H., Variational Methods for Eigenvalue Problems, 2nd ed. Toronto: University of Toronto Press, 1966. Hellwig, Giinther, Differential Operatorsof Mathematical Physics. Reading, Mass.: AddisonWesley, 1964. Hildebrand, F. B., Finite-Dzfference Equations and Simulations. Englewood Cliffs, N.J.: Prentice-Hall, 1968. Isaacson, E., and H. B. Keller, Analysis of Numerical Methods. New York: John Wiley & Sons, 1966. Iserles, Arieh,A First Course in the NumericalAnalysis of D~flerentialEquations. Cambridge: Cambridge University Press, 1996. John, F., Lectures on Advanced Numerical Analysis. New York: Gordon and Breach, 1967. John, F., Partial Differential Equations, 4th ed. New York: Springer, 1982. Kamke, E. D~yerentialgleichungenreeller Funktionen. Akademische Verlagsgesellschaft, 1950. Kantorovich, L. V., and V. I. Krylov, Approximate Methods of Higher Analysis, 4th ed. (trans. by C. D. Benster). Groningen: P. Noordhoff Ltd., 1964. K h h , T. V., and M. A. Biot, Mathematical Methods in Engineering. New York: McGrawHill, 1940. Kellogg, 0.D., Foundations of Potential Theory. Berlin: Springer, 1967. Lord Rayleigh, The Theory of Sound, 2 vols., 2nd ed. New York: Dover, 1945. Mikhlin, S. G., Integral Equations (trans. by A. H. Armstrong), 2nd ed. New York: Pergamon Press, 1964. Sneddon, I. N., Fourier Transforms. New York: McGraw-Hill, 1951. Sommerfeld, A., Partial Differential Equations in Physics (trans. by E. G. Straus). New York: Academic Press, 1964.

1

712

Advanced Calculus, Fifth Edition Southwell, R. V., Relaxation Methods in Engineering Science. Oxford: Oxford University Press, 1946. Southwell, R. V., Relaxation Methods in Theoretical Physics. Oxford: Oxford University Press, 1946. Tarnarkin,J. D., and W. Feller, Partial Differential Equations.Mimeographed notes of lectures at Brown University, 1941. Titchmarsh, E. C., Eigenfunction Expansions Associated with Second-Order Differential Equations, 2nd ed. Oxford: Oxford University Press, 1962. Titchmarsh, E. C., Fourier Integrals. Oxford: Oxford University Press, 1948. Wiener, N., The Fourier Integral. New York: Dover, 195 1.

714

Advanced Calculus, Fifth Edition

k) meaningless. -1

3. a) D - c = [ -

-4

-1

1:

2

b) G

)

=

[

r

-3

Section 1.8, page 25

I. a) meaningless; e)

'P)

[:

and

51

[i:

=5E;

c meness;

b) [I:];

1;

A;

I)

r) 1 2 5 g

q) 25E;

+

g) L;

+

3. a) yl = 17x1 x2, y2 = 3 1 x l 7 x 2 ; 6. It is true precisely when AB = B A . Section 1.9, page 30

+ I. a) 1 2 [ -2

-:I;

b)

A'[-!

-:I;

d)

3

3

1

3

[: I:] [

h) undefined;

and

-2

2 . 6],

i) 0; j) 0 ;

s) meaningless.

b) yl = x2

- 3 x 3 , y2 = 7 x 1 + 13x2 - 4x3.

735

Answers to Problems

8. All matrices whose inverses appear in the answers are assumed to be nonsingular. a) x = ;(A B), Y = ;(A - B); b) x = ( B - I)-~(BA - c),Y = (B - I ) - ~ ( c - A); C) X = B - A(A - c)-'(B - D), Y = (A - c)-'(B - D); d) x = (B-'A - E - ~ D ) - ~ ( B - ~-c E - ~ F ) ; Y= (A-'B - D - ' E ) - ~ ( A - ~ C- D-IF); e) X = (CB-I - FE-')(AB-' DE-')-~, Y = (CA-I - FD-')(BA-' - ED-l)-'.

+

-

Section 1.10, page 33 1. a) x = - 6 , y = 1 2 , ~ = 5 ; b) x = 3 , y = l , z = - 1 ; c) x = 13~ , y 10 = ~ , Tz9 = ~ . d) x = I , y = l , z = l . 2. a) 21; b) 0. 3. a) x = 2 , y = 2 , z = 3 ; b) x = 4 , y = l , z = l , w = l . -1 -1 -1 1 -3 4 1 4. a)

;

-2

-3

4353

3 -2 -2 b) no solution;

t

5. a) x = (7 - t)/5, y = (3t - 1)/5, z = t, -m < r < m ; C) x = - 3 t , y = 6t, z = t, w = 8 t , -00 < t < 00; d) x = t - 2 u , y = 5 t - 7 ~ , ~ = 3 t , w = 3 u , - m < t <~,-w:,u<<.

Section 1.11, page 38 Throughout, k is an arbitrary nonzero real scalar; c is an arbitrary nonzero complex scalar. 1. a) h = l , k ( l , - 2 ) , a n d h = 5 , k ( l , 2 ) ; b) h = O , k ( 3 , - 1 ) , a n d A = 7 , k ( l , 2 ) ; C) X = 1, k(O,2, 1);X = 2, k(l,2,0); l = 3, k(1, 1, -1); d) h = 1, k(3, -2, -2); h = - 1, k(-1, 1, 1);h = -2, k(2, - 1, -2).

C) h = 0, c(l,O, O), A = i , c(0,2, 1 - i), A = -i, c(0,2, 1

+ i),.

4. a) h = 1, all nonzero vectors; b) A = 0, all nonzero vectors; C) A = - 1, k(1, 1); d) A = 2, k(l,2,0) and A = 3, a ( l , 3 , 0 ) b(O,2, I), with a and b not both 0.

+

. l W r

3

>a

ti

, i;b

,

716

Advanced Calculus, Fifth Edition Section 1.13, page 44

[i 11;

4.

[i t]; [

1- a) b 2. a) a = 1; b) a = 4, b = 8. 5 2 3 1 ; b 1 ;

[

c

;+

- ?.

Q

c) c01(1,5,0.4);

[

:I;

3 1 -1

.

d

9

d) (1.0.7).

d

[

0 1 1 -2

i].

Section 1.15, page 53 1. a) (4,2,2,2), (8,6,2, -2), (6,4,2,0), (-3,0, -3, -6), (0,0,0,0) = 0; b) (7,6, 1, -41, (17, 12.5, -2), (-2, -2,0,2), (-6, -6,0,6); C) 4,24, 2. b) No; c) (6,2,-1,3,6); d) ( 1 0 , 0 , 8 , 7 , 4 ) , ( 1 3 , - 1 , 9 , 5 , 2 ) . 5. a) J38, COS-I COS-I (19/&73), c0s-l (19/&73).

a, Ji5.

a, a, (-A),

Section 1.16, page 60 1 a 2 3 1 5 3 1 1 2 b) Kernel is 0 alone, T is one-to-one, T(x) = (2,3) only for x = (1,O); c) Range is v 2 , T maps v2 onto v2. 2. a) (2,4), (3,6), (-1, -2), (-5, -10); b) Kernel is all t(3, -2), T is not one-to-one, T ( x ) = ( 2 , 4 ) f o r x = ( l , O ) + t ( 3 , - 2 ) , - m < t < m ; c) Rangeisallt(l,2);Tdoes not map v2 onto v2. 3. a) n = 3, rn = 2; b) all t(O,2, -1). not one-to-one, c) Rangeis v 2 ; T maps v3 onto v 2 . 4. a) n = 3,rn =2; b) All tl(-4, 1,O) t2(-3,0, l), not one-to-one; c) Range is all t(l,2), T does not map V' onto v2. 5. a) n = 2, rn = 3; b) All t(1, -2), not one-to-one; c) Range is all t ( l , 2 , -3), T does not map v2 onto v3. 6. a) n = 2, rn = 3; b) 0 alone, one-to-one; C) Range is all tl(2, 1 , 1) t2(1,2,2), T does not map v2 onto v3. 7. a) n = 3, rn = 3; b) All t(1, - 1, - I), not one-to-one; c) Range is all t1(3, 1 , s ) q(l,O, 2), T does not map v3 onto v3. 8. a) n = 3, rn = 3; b) 0 alone, one-to-one; c) Range is v3, T maps v3 onto v3. 9. Angle is r / 4 , 1x1 = IT(x)l. 10. Reflection in x-axis. 11. a) Reflection in origin; b) All vectors stretched in ratio 2 to 1.

+

+

+

Section 1.17, page 66 1. a) 2; b) 3; c) 3; d) 2 2. a) h = 1, basis (-10/13,4/13, 1); b) h = 0; c) h = 1, basis(-5, -22,6, 1); d) h = 3,basis(-l,2, l,O,O),(-1, l , O , l,O), (-3,4,0,0, 1). 3. a) (1,3,2), (0,0, -3,(0, 1, 1). or el, ez, e3 (W = v3); b 3 ,0 -142); C) (1,2,2,3),(0, 1, 1,2); d) (0, 1, -1,-3, 1),(1,0,2, 1,O). (O,O, 1, -2, 1). Section 1.18, page 70 1. a) 3, basis: 1, x, x2; b) infinite; c) infinite; d) 2, basis: ex, e - X ; e) 3, basis: Ell, E22, E33; f ) 10, basis: Ell, E22, E33, E 4 4 EI2 EZ1,E13 E31, El4 E41, E23 E32, E24 E42, E34 E43; g) 2, basis: cosx, sinx; h) 3, basis: 1, ex, e - I ; i) infinite; j) infinite; k) infinite; 1) infinite.

+

+

+

+

+

+

CHAPTER 2 Section 2.4, page 82 4. a) 0; b) no limit; c) 1; d) oo. 6. a) all (x, y), adomain; b) x2 + yZ > 1, a domain; c) x2 y2 i 1, a closed region; d) all (x, y, z) except the points of the xy-plane, an open set, not a domain.

+

'

Answers to Problems Section 2.6, page 89

zy = 2[1 - (x + 2y)2]-1fi; g) zx = ex(ez+ l)-l, zy = 2eY(ez + I)-'; h) zx = -(y + z)(x + 2z)-', z, = -(2xy + z2 + xz)(2yz + xy)-'. 2. Approximately fx(l, 1) = (2' -1 ( l ' = 3, or f x ( l , 1) = " 1 . 1 ) - 1 " 0 . 1 ) = 1, or fx(l, 1) = L w l p 9 = 2. It can be shown that the last value is "in general" the best estimate. For fy (1, 1) the analogous formulas give the estimates -2, -2, -2. This topic is discussed in Problem 9 following Section 6.18. = eYCOSY,a = eUCOSY; = -2; b) 3. a) = 2x,

(e)

(g)u

x Y

( 4

) = 1, .

)

( 2 )= x-d x;+ 1

b)*;

;

d) (X - 2y)4ex~[(xy- 2y2 f)

d)

= x(x2 + y2)-'12,

($)y

d

+ 5)dx + (x2 - 2xy - 10)dyl;

~ ) A (L+ A=X +~A ~, ) ~ 6. 2.2, 2.4,2.6.

z

=

1,

Section 2.7, page 95 5 2 2x2 . I. a 1 ; b) [4x1 3x2 3x1

[

cosy

= x sec 0 tan 0.

C) ( y - y (' )l d- xx+ -( ~ -) ~~2 ) d y .

5. a) Az = 4Ax

+

(&Iu

~

c)

e)

w, dx + x d

+ 2Ay + G2+ 2AxAy, dz = 4Ax + 2Ay; ( A x - A )(Ax + A ) , s o A zl D =+ ~ x~ + Az y ) -Y .

[;i:3

xr

-x sin y

0

3. a) 9(x2 + y2)2, b) 0; C) 2u3v2w; d) 4t2(x2 - y). 4. a) e2 = 7.39; b) 7.44; c) du = edx; dv = e dy, e2 = 7.39. Section 2.8, page 100 2. = Y u u - + u u o g . 4. 1. 5. 336. 6. 197.

3.

2

1

dv

717

718

Advanced Calculus, Fifth Edition

10. In all cases dz =

2 dx + 2 dy:

a ) d t = 2cot ( x Z y 2- 1)(xy2d x

+ x Z ydy);

:' '

.'

Section 2.9, page 104 1. a )

[

-

2u2

~ 0 ~ 3 x -23 x 1 s i n 3 x 2 ] , [ - 3 0 1 . sin 3x2 3x1 cos 3x2 2 0 '

4I. 1

+2

2u2

+ 2u1 - 1

Section 2.11, page 116 1. a ) 2 x / z , y / z ; b ) -(yz 4xz 3z2)l(xy 2x2 6xz), - z / ( y 2x 6z); C ) -zl(3z2 x 2y), -2z/(3z2 x 2y); d) -zexz/(xexz yeYz I), -zeyZ/(xexZ yeYz 1 ) 5 a 5 a 2. = -3, = - $. = -7, = 3. 3. a ) 3 x 1 ~Y./ U ; b) ( X U vy - ~ e " ) / ( e ~+ xeU ~ xeU - x 2 y2),(vev - x v - yu)/ (eU+"- xeU xeu - x2 y2). c) -(4xy 2yu u2 - 2yuv)/(2xy 2vy + x u - 2xyv), (2uv - 2yv2)/ (2xy 2vy xu - 2xyv). 4. a ) du = d(dx 3 d y 2 d z ) ,d v = - f ( d x 2 d z ) ; b ) ( & ) y , i = ( $ ) x , i = 0; C ) u = 3.033, v = 2.1.

+ +

+

+

+

+ + + + +

+

+ + +

(g), (t), (g), +

+

5. (3yu - 4xu

+

+ 4x2 + 9xy + 8 x v ) - ' [ ;

+ +

+

(g)x i +

+

6,

+

+

+

i ] ,where

+

o = 8uv

.X (b

+

+ 4u2 - 8xy - 4xu.

b = -3u2 - 8x2 - ~ X U c= , 4xy - 3y2 - 6yu - 4xv - 9yv - 4uv - 8v2, d = 4x2 - 3xy 6xu 3uv. 6. a ) -1; b 1 ; c ) -1, -1.

+

+

i,

Section 2.12, page 121 2. a ) u = ~1 ( x + ~ ~ ) , 1v = ~ ( Y - ~ b)x J)=;5 , J f o r i n v e r s e = $ . 3. a ) J =4(u2 u2), b) ( Eau) y = 2(U2; y 2 ) . ( Eav) y = - 2 ( y 2 ; U 2 ) . 6. a ) J = ,02sin,?

+

b)

% = s i n # s i n ~ , # ---

Section 2.13, page 127

, 80 E =-=.sin 0 b

1. b ) x = $ + i ( t - ~ ) , y = i - p ( t - g ) , z = $ + Q ( t - 5 ) ; x - Ay &z = 3&/4. 2. b ) f i x 2y z = 1314. 5. b) 3 & i x - y - 4 z = l . 8. a) 2 ~ + 2 y + ~ = 9 , + = + = Z - ' .

C)

+

i

(0

iF;

.&

+ +

1

'

719

Answers to Problems

=+4k

5 %; + + q;

1 b) z = 1,x = ~ , =O; y c) 2 x - y - z = 0 , 9 = = d) none; e) ylx xly - z = x ~ y ~f ;) (yi z ~ ) x (XI ZI)Y (XI YI)Z= 2. 10. a) z - 2 = 2 ( x - 1)+2(y - 11, = Y_?_1 =

+

+

+ +

+

2

9

.de

z-L

b) z - $ = -2(x- $) -2(y - $), = = +; c) z = x - 2 y + 2 , x = 2 + t , y = 1 - 2 t , z = 2 - t ; d) 5 ~ = 6 ~ + 8 y - l O , x = $ + g t , y =4~ + 8j t , z = - t . 11. a) x = 3 + 4 t , y = 1 - 5 t , z = 1 + 3 t ; b) x = 2 + t , y = 2 - t , z = l ; c) x = I , =~ - l + t , z = - t .

Section 2.14, page 134 1. a) b) 0; c) ;

d)

-5; e -4;

f ) 0. 2. a) x2

- Y';

7xyz 6. dulds = -2xy. Minimum is -4 at (0, f2). b, ,/x2 +4y2 + L6z2. 8. Maximum is 6 in direction i at (1, 0), in direction -i at (-1,O). 9. 3,

Section 2.18. w e e 142

~fll

9

1 l5 43' 33' 1.

Section 2.21, page 158 1. a) maw. at -1, min. at 1; b) max. at n/3 2nn, min. at -n/3 2nn, horiz. infl. at n 2nn (n = 0, f 1, f2, . . .); c) max. at log 2. :! , 2. min. for n = 2,4,6, . . ., horiz. infl. for n = 3,5,7, . . . . 3. a) max.=l,min.=O; b) max.=O; c) nomax.ormin.; d) maw. = 21 , min. = - 21. 4. a) max. at (0,O); b) min. at (0,O); c) saddle point at (1, 1); d) saddle point at (0,O); e) critical point at every point of the , , ,,, line y = x, each point giving a relative min.; f) triple saddle point at (0,O); . , g) min. at (A, 2nn), (-&, 2nn), neither at (0, nn. - 2 ) ; h) min. at saddle points at (0, O), (1, O), (0, 1); i) (0,0), neither max. nor min.; j) min. at (0,O); k) min. at (0, fI), saddle point at (0,O). 5. a) abs. max. at (0,O); b) saddle point at (0,O);

+

+

2+

(4,3).

9+

+

720

Advanced Calculus, Fifth Edition .

+

-

c) saddle points at n / 2 2n1r (n = 0, f1, f 2, . . .); d) no critical points [discontinuity at (0, O)]; e) abs. min. at (0,O); f ) abs. max. at (fi/3, &/3). 6. a) abs. min. at (-315, -419, abs. max. at (315,415); b) abs. rnax. at ( f 2-'I4, f 2-'I4) (four points), abs. min. at (f 1,O) and (0, f1); 1 1 C) max. at (f3, f 4), min. at (f4, ~ 3 ) ; d) max. at (L, 0, -), min. at (---,1 0, - -); .h .h .h .I3 e) m a . at ( and (0, - 1.0). min.at ( f f and (0, 1,O).

&,-A, &)

* &.&.*fi)

f ) Abs. min. at (1 - &/2, 1 - &/2, - I), rel. min. at (1 &/2, 1 &/2, -& - I), no abs. max. 7. (0, f 1,O) and ( f 1, 0.0). 8. a) m a x . = l ; b) m a x . = 21, m i n . = - 2 ; 1 c) max.=3,min.=-3; d) m a x . = l .

+

+

1 9. (a) and (c) are positive definite. 11. a = m(2el - e2 - 2e3 - e4 + 2e5), 1 b = rn(-2el - e2 e4 2e5), c = &(-3el 12e2 17e3 12e4 - 3e5). 12. &/3.

+ +

+

+

+

Section 2.22, page 166 1. 1,0, sinx. 3. a x by c, a, b, c arbitrary constants. 5. f (x, y) = g(x) h(y), g(x) and h(y) being "arbitrary functions."

+

+ +

CHAPTER 3 Section 3.3, page 180 6xy 3x2

2 4

0

10

Section 3.6, page 185 5. a) X ~ ~const; Z b) (2y + z2)eXy+const. 7. a) yzi - 2xzj+ grad f,f arbitrary; b) i[z2i (x2 - 2yz)jl +grad f,f arbitrary. 12. b) c) 2x2i 2y2j. 15. 16. d i v v = O , c u r l v = 2 w . 17. Vol.=l. 18. Vol.=e.

-3.

+

+

g;

i ;

+

Section 3.11, page 210 1. a) ul = -36x1x1 51x1x2 - 18x2x2 - 28x1 20x2, u2 = 24x1x1 - 34x1x2 12x2x2+ 21x1 - 15x2; b) v' = (3x' - 2x2)[3 cos (-4x1 3x2) 2 sin (-4x1 3x2)], 3x2) 3 sin (-4x1 3x2)]; v2 = (3x1 - 2x2)[(4 cos (-4x1 C) wll = ~ 2 = 2 0, ~ 1 =2 -w21 = -12x1x1 + 17x'x2 - 6x2x2; 'I d) = -71x1 49x2, = 53x1 - 36x2, z: =,-103x' 72x2, zi = 77x1 - 53x2. 2. In standard coordinates, all reduce to & . ,In (x' ), gll = g22 = 25, x i W g12 = g21 = -g12 = -g2' = -18, g22 = gl' = 13, g; = 8ij always. 3. a) u' = 13ul 1 8 ~ 2u2 , = 18u1 25u2, where u ~u2, are as in the answer to Problem l(a); b) vl = 25v' - 18v2,v2 = -18v' 13v2, where v l , v2 are as in the answer to Problem l(b); c) wll = w22 = 0, wI2 = -w21 = wlz, where wl2 is as in the answer to Problem l(c).

+ +

+

-+

Z!

+

+

+

+

+

+

+

zi

+

+

+

CHAPTER 4 Section 4.1, page 219 1. a) 2x sinx - (x2 - 2) cos x

- log (1 d) (n/4)+log(l/&). d) 2(-

+ -))

+ c; b) ;arc tanx2 + c; c) log % + C; + C. 2. a) n/4; b) 0; c) 3e - 8;

3. a) x12;

b) 1;

c -1

d) log(l+&);

e) 2;

i

f ) 1.

Answers to Problems

721

b ) div; c) n / 2 ; d ) div; e) div; f ) log2. 5. a) 413; b ) 1 ; 14-n-6fi. 6. a) 2 / n ; b ) - 2 / x ; c) 112, d ) b + $ a ( x l + x 2 ) . 8. a) value 0.3095, error at most 0.0082 (worst error at x = 1); b ) value 0.7667, error at most 0.1321 (worst error at x = 1). 4. a) 0; C)

Section 4.5, page 234 1 a) 1 3 b ) n/48; c) (-15&)/8; d ) 1. 2. a) e - 1; b ) (2 - 5e-')(l- e-2); 2 I+x c) 314; d) n/6. 3. a) f l f1_ f dy dx or f dx dy, where )(y) = I1 - yl f o r - 1 ~ y ~ O a n d f o r 2 ~ y j 3 . # ( y ) = l f o r O ~ y j b2); f i J ~ ~ ~ ; : , f dory d x

/Il /t(y,

J!{:, /:,!:$:

f dx dy, where @(x)= ( x - x2)'fl. +(y) = $ ( l - 4y2)112.

4. a) ( 8 / 3 5 ) ( 9 f i- 8&

+ 1);

b ) 215. 5. a). fo

112

1-y flI2 f dx dy;

f j hm

f dx dy .

f dx dy; c) f 1 1;'' f dy dx; d ) lo. a) i i - (e - 1 ) j + log 2k; b ) &(i + j).

b)

Section 4.6, page 241 1. a) An; b ) 2& - 2

+ 210g (2&

- 2);

c) &(tan-'

& - tan-] 1 A );

d ) logc + 1 - c, c = n/(4&). 3. b ) cosh 1 - 1; c) log 2. 4. a) n / 2 ; b ) (14& - 7119; c) n4/3; d ) 1 1 - 1 log T3 ; e) for 0 j u j 1, f ( u ) = 0 for 1 j u 5 e, 5. + :2u2 du du. where f ( u ) = andg(u)=-for0ju se. 7. a) 2 jO4I:-'' log ( 1 2u2 2v2)dudv; b ) fd f,l-2u ,/I u2(u+ v ) dv ~ du. 11. a) 1 : f; fd r4 cos28 sin 8 dz dr do; b ) f:/2 fd f~+r(COSe+Sine) r cos 28 dz dr do. 12.a) Jb2"f~f~p5sin4#cos28sin8dpd#d8;b ) &2n fon/4 fosece p 4 sin)dpd#de.

3.

f i 17:;;

+

+

+

.

Section 4.7, page 248 3. a) 4n2ab; b) n(s3I2- 1)/24.

=k

.

1

.

Section 4.8, page 252 2. -n/2. 3. a) 4n/(3 - p); b ) 4 n / ( p - 3). 4. a) div; b ) conv; c) div; d ) conv; e) conv. Section 4.9, page 256 1. a) sin ( x t )d x ; b

fG2

j12

2.$

d

c) f:

d x ; d ) n!1:

dx.

+

2. a) x2; b) 2t sin t4; c) -3t2 log (1 t6); d ) sec2xe-'8"2x - e - ~ ~ . nl n 1 . 3 ... (2n-3) 1 4. a) (n+1) b) C) + ~ . 4 , . . ( 2 , , - 2 ) p i r r ? ~ > O .

3;

Section 4.11, page 264 1. b ) uniformly continuous. 2. All but b ) are uniformly continuous. 4. a) f must be uniformly continuous; b ) f need not be uniformly continuous.

CHAPTER 5 Section 5.3, page 278 1 a ; b c) 0. 2. a) $; b ) 8; C ) -n/2. 3. a) 0; b ) -2n; C ) 4. a) 0; b ) a / 2 ; c) 6. a) 7; b ) 5; c ) 8; d ) 5.5; e) 0.

-4.

1 8+ 4 log (2 + 3).

*-@

Advanced Calculus, Fifth Edition Section 5.5, page 286 1. a ) b) c ) $. 2. a ) O; b) c ) $. 5. a ) ( b - a ) times area enclosed by C ; b) 0 ; c ) ;n; g) 0 ; h) 0.

4;

.

i;

i;

d ) 0;

+

e ) 0; f ) 47r;

Section 5.7, page 300 I. a ) F = x Z y ,integral = 1; b ) F = eXy,integral = O; c ) F = - l / J W , integral=1-e-2". 2. a ) b) C ) 2; d ) 1. 3. a ) 0 ; b) -2n; c ) -2; d ) 1217; e ) 0 ; f ) 0. 4. $ + 2 n n ( n = O , f l , f 2 ,...). 5. a ) 0 ; b ) 217; c ) 417; d ) 0. - nkak, where 6. a\ x 2 y - ;(y3 2); b) x sin y. 7. -n. 8. K nlul n22. the numbers n l , . . . , nk are positive or negative integers or 0 9. 7. 10. a ) -60; b ) 0.

-9; 4;

+

+

Section 5.10, page 312 5 1. a ) 3 n ; b) -T; c) 7. a ) n / 4 8 ; b ( e - 1

+

+ +

i; 3nd ) 0 ; -(

e ) -163115. 5. a ) $; b) n ; c) 2 n ; d ) e -1 c ) n / 8 ; d ) (e2 - 1)/2; e ) 0.

5.

2

Section 5.11, page 319 l . a ) 4 n ; b ) 3 ; c ) O ; d ) 6 V ; e)O; Section 5.13, page 330 1. a ) 617; b ) 0. 2. a ) -2; b) 0. 4. where v = x2y2zi - xy3zj + xy2.z2k.

f)O.

5f

2nn. 5. b) v

+ grad f,

Section 5.14, page 336 1. a ) ji j;(u2 v 2 ) d v d u ; b ) jjR,,;(u - v)(1 - 2u - 2v)du d v ; C ) 4 jjR,,,uu(u4 - v4)du dv. 3. a ) -1; b) 2; C ) 3. 6. a ) -1;

+

b ) 2.

Section 5.15, page 347 LZC. Xl.lc.. 1. a ) Potential energy is ;k2x2; ;mu2 $k2x2= const. b ) Potential energy is 1 i ( a 2 x 2 b Z y 2 )$(mu2 ; a2x2+ b2y2)= const. 5. T = TI + 9 . x .

+

+

+

Section 5.18, page 363 3. (a) and (b)2nab log l / a for R c a , 2nab log 1 / R for R > a ; c ) O f o r R < a , - 2 n a b ~ - ~ ~ f o r R >4.a .a ) - 2 n y f o r R < a , O f o r R > a ; b) normal derivatives have limits 0 everywhere. 6. U = nk(4a3 - R 3 ) / 3for for R 1: a. R 5 a, U =n k a 4 / ~

*?a

Y

'a,...

.S

r

Section 5.22, page 373 1 1. a ) x1(x4)' dx' - (x112x4dx4; b) -xl dx2 d x 3 - x2 d x 3 d x l - x3 d x l d x 2 ; C ) 2x1x4d x l d x 2 d x 3 - 2 ( ~ ld x ) 2~d x 3 dx4 + 2 x 1 x 2d x l d x 3 dx4 - 2x1x3d x l d x 2 dx4; d ) 0 (4-form); e ) x4 dx' x 1 dx4; f ) 0 ; g) 3 d x l dx2 d x 3 . 5. a ) ( f + 3 f 4 )d f d f 2 + ( 6 f 4- i 2d) . f l d . f 3 ( - 3 f 2 - 6 f 3 )d f l d f 4 ( f l + 6 f 4 )d f 2d . f 3+ ( 3 f 1- 6 f 3 )d f 2d f 4 ( 6 f 1 6 f ' ) d f 3d f 4 ; "q b) d f l d i 2 - (.i1)3f2 d f l d z 3 ( z ' )d~f 2d f 3 . d . f 2 X3 d z 3 on 6. a ) j ( X I X2 X3)d f b) j j X I d f d f ( X 2 - XI)d i 3 d f l ( X 3 - X 1 ) d i ld z 2 on 3. 8. C,jk = AIBjk AkBlj AjBkr. t ' i J

+

(R')~Z

+ + +

+

+

+

+

+

+

+

+

+

+

c;

C*

A.-

dii Answers to Problems

CHAPTER 6 Section 6.4, page 383 , b) 0.951, -0.951; 1. a) 0; b) 0; c) 0; d) 1; e) 1 2. a) 1 c) oo,-oo. 6. Upperlimitisooforix( > 1 , l f o r x = f l a n d O f o r ( x l < 1;lowerlimit is-ooforx < -1,-lforx=-1,Ofor-1 < x < 1 , 1 f o r x = 1,ooforx z 1. Section 6.7, page 396 12. a) conv; b) conv; c) div; d) conv; e) conv; f ) conv; g) div; h) conv; i) div; j) conv. 14. a) div; b) conv; c) div; d) conv. Section 6.9, page 404 T' 1. a) 1 term, 1; b) 3 terms, 0.86; c) 5 terms, 0.51; d) 2 terms, 0.70; e) 3 terms, 1.287; f ) 4 terms, 1.708; g) 3 terms, 0.8417; h) 1 term, 0.72; i) 1 term, 0.18; j) 8 terms, 1.70. 2. a) 8 terms, 66 terms, 918 terms.

7r2

5n2

7r4

a4

b) a + w ; c) 7 - m ; d) -6+,7q,+TiSJ, -. 7. b) 0.9729; c) 1.0766.

1- a) n 2 ; f

n4

=

, ,

Section 6.10, page 409 7r2

P

n6.

n2

n6

15.

e) T - W - & i .

(1

Section 6.13, page 417 l a x 1 b) I x l < 2 ; c ) x > l a n d x < - 1 ; d)x>O; e ) x < $ ; f) I x - n n ) s n / 6 ( n = O , f l , f 2 , . . . ) ; g) 0 1 x 5 2 ; h) x > l a n d x < - 1 ; i) all x; j) all x. Section 6.16, page 429 3. a) ~ ~ o ( - l ) n ( x 0 < x < 2; b) E ~ ~ ( - I ) " & . -2 < x < 2; (-1)"3" 5 5 (- 1)"3" 5 11 c) C z 0 T ~ n , - 3 < X < 5; d) CEO (x< X< 5; e)

xzo(ac+b)fl+~(x - c ) n 7 c - 1-aczbI i x
k(k+ 1) . . . k+n - 1) an(x - c)" ac+b 1 j) ( a ~ f b ) ~ 1 .(.. n (a,+b)n+k*c< 4. b) f(0) = 0, f(1) = 1.29, ff(0) = 1, f f ( l ) = 1.63, f"(0) = 4; rn n+l +

1 7 1

Crn


Section 6.18, page 436 4. ~ x , 2 . a ) x + x 2 + 3 x 3 ; b ) x + f x 3 + & x 5 ; c ) x ~ - x 11 ~ + d) -x2 - $x4 - ;x6; e) 15 15(x - 2) 6(x - 2)2; f ) 1 x i n 2 ; g) x - ix3 + &x5; h) x - 4x3 Ax5; i) x 4x3 ;x5; j) ;x2 + Ax4 + &x6. 5. a) 0.747; b) 0.497.

+

+

+

+

+

+ +

1

ac+b 1 7 1 '

724

Advanced Calculus, Fifth Edition Section 6.19, page 442 1. a) 0; b) 1 - i. 2. a) absolutely convergent; b) divergent; C) convergent, not absolutely; d) absolutely convergent. Section 6.21, page 447 1. a) 1 (x2 - y 2 ) &(x4 - 2~~~~ y4)

+

+

+ + - .. + $(x2 - y2)n+ . .. , all (x, y);

n-1

(X )2n-'

b)xy-hx3y3+...+(-1) h+-.-,all(x,y); c) 1 + ( x + y ) + ( x 2 + 2 x y + y 2 ) + . . . + ( x + y ) " + . . . , - I < x + y < 1; d) ~ + ( X + ~ + Z ) + ( X + ~ + Z ) ~ + - ~ ~ + ( X + ~ + Z ) ~ + ~ <~ X ~ ,+- Y ~ + Z< 1. 4. 0.240 (three significant figures). Section 6.25, page 460 1. a) divergent; b) convergent; c) convergent; d) convergent for s > 0; e) convergent. 12. a), c), d), e), g), h) principal value 0, no usual value; b) both values 0; f ) both values (312) (52/3- 1).

CHAPTER 7 Section 7.4, page 478 1. a) ~ + 2, ( s i n x + ~ + ~ + . . . ) ; A

b)

$-

cos 2n - (2n - ~ -m1)

n=l ( - 1 ) n nh . 9 d) $+4Ez1-;

f)

3. a) x - 2 C E 1 F-$cm ,,=I 7 cos nx .

4. a)

F;

b) 6n;

c) 2;

e) + x F l nh .7

-y-, sinnx .

d) not periodic;

b)

e) 2n;

$ + i x ww c o s 2 n x . n=1 4,,

f ) 30n;

-1

g) not periodic.

Section 7.5, page 484 l . a ) 1 - 4 ~ ~ ~ ~ s i nb)n 1x+ 2; n - 4 C Z l Tsin, nx . 2 m 1 - (-1)"(2n + 1) C) 1 + IT. - 8 ~m cos (2n - Ox. K n=l (2n * )' R En=] n l - 2 . 1 . 2. a) ~ + 4 ~ ~ ~ l ~ c o s n ~ - 8 n ~ ~ f c o s ~ n r- 2 m sin2nxx x Cn=l- n . 2 4 ca cos(2n - i)x; 3. a) sinx; b) and c) ; r - T4 i xm, = cos2nx lz. 4. a) (2n-l) e) +

b)

:

+

d) - i z z z l ( - l ) n sinnnx b) -2Cz,(-1)"n ; C) 2 1 - $,=I 4 E@ (2n!A1) p; 5. a) bn = 0; b) a, = 0; c) a2n+l = 0, bZn= 0; e) bn =O;an =Oexceptfor d) a4,+2 =O,b4,, = 0 , ~ 4 ~ +=1b4n+l.~4n+3 = -b4,+3; n = 0 , 4 , 8 , . . . , 4 k , . . .; f ) a2, = 0, b2,+1 = 0; g) azn+l = 0, b2,+] = 0; h) a, = 0 and bn = 0 except for n = 0,4,8, . . . ,4k, . . . ; i) an = 0 and b, = 0 except for n = 0 , 6 , 12, . . . , 6 k , . . .; j) a, =Oandb, =Oforn = 1 , 2,.... nn

Section 7.13, page 506 13. a) cos 2nt, 0 5 r 5 n/2; b) (-1)"12 cos nt for n = 0,2, . . . , (-I)("-')I2 sin nt for n = 1,3, . . . , -n/2 5 t 5 n/2; c) (-1)"12 sinnt, for n = 2,4, . . . , (-1)(n-1)/2 cosnt for n = 1,3, . . . , -n/2 5 t 5 n/2.

An-

to Problems

15. a ) ( 1 - t2)-'I4 cos (n cos-' t ) = Tn(t)(l- t2)-'I4; b ) ( 1 - t2)-'I4 sin (n cos-' t ) = U,(t)(l - t2)'I4; C ) ( 1 - t2)-lI4 sin [(2n+ 1 ) cos-' t ] = Uz,+l(t)(l - t2)''4; d ) ( 1 - t2)-'I4 cos [(2n 1 ) COS-' t ] = Th+l(t)(l - t2)-1/4.

+

Section 7.19, page 523

&

$+

* *

3

+

cos m

1 3 m ~ ( - l ) n y %$ ~:=~(-l) + 1 6 C z 1 ~ ~ = = , ( n -m I ) ~ + " ~ . 1. b)

b) 1 100sin r cosxt + (1 - cos r) sin xr

2. a ) 1 JW c o s x r + t s i n x r

t. n o t 4. a) (2sint)lt; b ) ( 1 +it)-'; c) (2t-' - 4t-3)sint + 4 t - 2 ~ t. ~ ~ 6. b) for o > 0 , (2n)-' k!e("+")+(o it)-k-' d t = xk for x > 0, = 0 for x < 0. 7. a) h(x) = x ; b) h(x) = x2/2; C ) h(x) = eX - 1. l+r2

n 0

+

[-%

Section 7.20, page 529 3. a) p(x) q ( x ) 156(x) - 26(x rr) + 26(x - n ) , where p(x) = -1 for x > 0, q ( x ) = - sinx, -n 5 x 5 n , p = q = 0 otherwise; b ) 36'(x) - 6(x); C ) 26(x - 1); d ) 6'(x); e) -f (x)g(x) (3 - n)6'(x - n ) 6(x - n); h) 3s - 1 ; i) 2t3 sinnt/(l - t2)(valuen3fort = &I); f ) 1; g -1; j) p(x) 36(x), with p(x) as in answer to (a); k) 96(x) r(x), where r(x) = x - 6 for :d x 2 0,r = 0 otherwise.

+

+

+

+

+

+

+

CHAPTER 8 Section 8.2, page 536 5. a) 1 - i b ) 4t3 2it; c) 3ie3"; d ) (-2 + 4i)e(-'+2)'; ), e(-1+21)t[ l t(-1 2i)l; f ) -e-' + eit. 6. w' = 6ie2", w" = -12e2". 7. a) (3 i)/2; b ) (3 20i)/15; c) (e3' - 1)/3i; d ) 2(e-'+" - I)/(-1 e) [ l (2i - 2)e-'+2']/(-3 - 4i); f ) 2 - e-I - ei. 8. a) (t2 - 2it - 3) cost (2i - 2t) sin r c; b ) -e-af (f + + . . . + $) + c;

+

+ +

+

+

9

+ +

+

n(n -b3l ) ~ " --~ n(n - l ) ( n -b5 2)(n - 3 ) P 4 ~)m~bt[-$+ +

sin bt[nrn;' b

d) R~{-c"[%

n(n - l ) ( n - 2 ) P 3

< b4

+ (-all

-1.. .

+

+ 2i);

+...I+

+...I

C;

+ A]] + c; (-ai)*+'

+)]+c, e) -izZ, [e-'kt($ where the ak are the eight numbers (fa f b f c)i. + . . a +

Section 8.4, page 540 1. a) u = x2 - y2 - 2xy, v = x2 - y2 2xy, all z; b ) u = ( x 2 y2 y)[x2 ( y I ) ~ ] - ' , v = -x[x2 ( y I ) ~ ] - ' ( Z # -i); c) u = tan x sech2y [ 1 + tan2x tanh2y ] - l , v = tanh y sec2 x [ l + tan2x tanh2y]-' , z # ( ~ r / 2 ) + n n , n = O , f l , f 2,... ; d) u = e-X[(l+ X ) C O S ~ - y siny][(l+ x ) + ~ y2]-', v = -e-'[(l + x ) s i n y ycosy][(l+ X I ' + y2]-' ( z # - 1 ) ; e) . . .i) See (8.13),continuous for all z; j) u = eX(cosx cos y cosh y sin x sin y sinh y), v = eX(cosx sin y cosh y - sin x cos y sinh y), all z.

+ +

+ +

+

+

+ +

+

726

Advhhced Calculus, Fifth Edition

Section 8.5, page 544 1. a) b) n 2 / 2 ; C)

-l

i log2+i(n/4).

3;

3. a) 2ni

Section 8.6, page 549 2. a) Analytic nowhere; b) analytic nowhere;

c) analytic for all z. 6. The functions are analytic except at the following points: a) n n ; b) n n ; c) I n i n n i ; d) 0; e) 0, n n , where n = 0, f1, f2 , . . . . nn; f ) Section 8.7, page 552 1. a) 0.693 2nni; b) i ( $ n

+

d) exp (-

i n - 2nn);

in+

-in +

in+

+

+ 2nn);

c) 0.347

+

+ i ( $ n + 2nn); I) exp ( g n i + 2JZnni);

e) *exp ( b n i y i ) ; h) 2nn f 1.317i; g) 2nn; The range o f n is 0, f 1, f2, . . . , except in (e),where it is 0, 1,2. 3. a) and b) n n and ( n / 2 ) n n (n = 0, f1, f 2, . . .). 6. a) l o g r + i B , i n + 2 n n < B < ; n + 2 n n ( n = O , f l , f 2 ,...); b) f i e x p (i8/3),-(n/2) 2nn < 8 < ( n / 2 ) 2nn (n = 0 , 1,2).

in +

+ +

+

Section 8.9, page 558 1. a) 0; b) 0; C) (2i - l)e2i; d) -i. 2. a) x i ; 3. a) a1 a2 a3; b) a1 a2; C) a2 + a3. 5- a) 6ni; b) -2ni/3; c) 2ni; d) 2nisinhl.

b) x i .

+

+ +

?

Section 8.11, page 565 1- a) 1 b) 3; C) 0; 4- a)

C)

(- 1)"z2"+'

Czo

I. log2 + :ni

d) W . 3. a) 4nei/3; b) - 3 ; c) ni/32. (r* = R = w); b) C z o ( - l ) n ( z - 2)n (r* = R = 1 ) ;

-Czl

(v)"

QLL!~!! (*, = JZ, R = 1).

Section 8.14, page 574 1 1. a) removable; b) - g

$,

-

- Ego

+

e cos 1 +sin 1).

d) ! + ~ + & + f . . , p o l e o f o r d e r l .

c) -!+l+~z,.$#$,~oleoforderl; 1

+7+7;

2. a) ; i

3. a)

00

3

1

1 -

2

&,pole of order 2;

1

1

-1 7 .removable. zero o f first order; b) z - 2 +

+ xzl 5 +

5,

1 1 d) ;r - i.

xz, 9, pole of order 1;

C) 2 C z l essential. 5. a) zero: 0, pole: 1; b) zeros:l,oo,poles:-1,-2; c) zeros:-1,-1,-l,poles:O,oo,oo. b~ 6. azbfi - ~ I ~ N ~- aNo +b ~I + z +sob:+, b i ( z - zo)N-2

Section 8.18, page 586 1. a b) n i sin6; c) -ni/3; 1 1 1 1 h) 4ni. 2. a) Lfx-Lfn; b) C)

-:[&+&

+&+

d) 0; e) - x i / &

1 3 =--+-;

&+&].zk

f ) 0;

g) 4ni;

2

=exp[(2k- l ) n i / S ] , k = 1 ,..., 5;

727

Answers to Problems 1 . i&-4(z+l)2. (8

d) ~n [Z -~Z I + . . . + & - ] , z ~ = e x p ( ~ ) , k ,.... = l n; el + (-~)"-'co 1 f) + + . .+ c n - I - c n - 2 + z (-l)n4(-l)m; 1 1 5 1 7 1 go

3 9

g)

+

+n

- r-

~,~-,).7

x+6=1

+ +

i-z,9

where zl , . . . ,z4 are the roots of z4 z 1 = 0 and g = (-7z3 1 -2-i -2+i 2+i h) m [ T = T + ~ + * + z + l + i l . 7. 2A'/Bn. Section 8.19, page 590

1 a ;

b) $ n ;

3. a) ine-';

c)

3; 7rfi. 7T

b) - 2 q n e - f i s i n l ;

2. a)

q;b) 9; c) $;

c) !jne-:.licos

Section 8.21, page 1600 7. a) none, b) 2 n + n n ( n = O , f l ,...), z-1 (ii) W = f . 9. d) ( i ) w = (l-2i)z-l,

.B

+ 5z2 + 3z - 13)-';

(!jJr); d)

d) f .

-$nc-'. : .T

i

c) f l .

. Section 8.23, page 609 1 z i -1 b) $(IT-argz4), c) 1 - 2 1. a) ,arg-, (O<9 <2n), z +1 .I# 2 Xz-l 2 z-1 2 d) 2 - , a g ,K+!, ee) l - z a r g ( r n ) . The argument funct~onsare all taken between 0 and IT. arg z, where z is the inverse of the function 2. b) U = w = $ ( 2 ~ o g z - z ~ + 1 ) , Im(z) > 0; 3. Tl n arg ( d m- i); the square root is chosen to have imaginary part > 1 t and the arg 1s chosen between 0 and n. 5. b) [ahj (n - a ) h j + ~ ] / n . J

\

'

+

+

Section 8.26, page 617 i log z (not single-valued). 3. To 2. F(z) = z

+ +

+ x(Tl - TO). 5

Section 8.27, page 622 t~ 1. a) The domain bounded by the lines u = f n and the ray u = 0, log 4 5 u < CQ. b) the domain bounded by the lines u = n , u = -2n and the rays: u = -n, 1.89 < u < CQ;U= 0 , 1.28 5 u < CQ;V= 0 , -00 < u 5 3.17; c) the upper half of the w plane minus the portions of the imaginary axis between 0 and 0.27i and between 3.7i and CQ. 2. a) ;r1 2ni - 2 Log (z - 1) - log 41; b) $ [ ~ o g & + 2 ~ o g + + n i - Log (z - 5.1) - 0.031; c) (1 i)z-'I2(z - 1)'14.

LO^ 9 +

+

CHAPTER 9 Section 9.3, page 628 1. a) k(e2X- x2) C; b) c ~ x c2; C) &x4 c1x2 czx ~ 3 ; d) clxn-I C Z X , - ~ + . . . +c,-~x+c,; e) $ + c l x n - l + . . . + c , ; f ) l o g \ x l + c ( x # O ) . 2.a) 2-cosx; b ) e X - e x + l ; c) ex. 3. a) ; ( x 2 + x ) + l ; b) 1. 5. Order: a) 1; b) 2; c) 1; d) 2.Degree:

+

+

+

+

+

+

j

728

Advanced Calculus, Fifth Edition

Section 9.4, page 633 5. a), b ) , d ) are linearly dependent; c) is a linearly independent set. 6. a) cleh ~ z e - ~ b) ; cl ~ ~ e C- ) cleX ~ ~ ;c2xeX c3x2eX; d ) ef Jix[cl cos ( : f i x ) +c2 sin ( : f i x ) ] +e-ifix[c3 cos ( ; f i x ) sin ( $ f i x ) ] ; e) C I c2 cos x c3 sinx c4x cos x csx sinx; f ) ~ ~ e - ~ ~ ; g) e-ft [ C I cos (13& t ) c2 sin ( 1 3 f i t ) l ; h) cl C ~ X c3x2 c4x3 + c5x4. 7. a) cos x ; b ) l e - f t sinh ( w t ) . 8. a) - - f ( e 3 ~- e5-2x); A3 I-e b) cosx +2sinx; C ) csinx. 9. a) clex + c2e-X i x e ~ ; b ) cleX+ c2e2x + c3e3x + i e 4 ~ ; c) clcosx +czsinx - sinx - sinxlog(cscx +cotx); d ) cl cos 2x c2 sin 2x f cos 2x log 1 cos 2x1 + i x sin 2x; e) cleX c2e-' ;ex J e-' logx d x J e X logx dx. 11. clx c2x2 i n 3 . 12. cleX - $ ( x 2). 13. a) cl c2e3X c3e-3x; b ) cl c2x c3x2 c4ex ~ s e - ~ C; ) -ex + ~~e~~ c2e-3X. d) - 1 clex c2xex c3x2ex. 14. a) - i x cos x ; b e c) -x2 - 2x - 4; d ) -&(4cos 2x 3 sin 2x) $ex; e) ex(5x - 2); f ) & x e - ~ ( 3cosx - sinx); g) x3e-'. 16. a) (f )(e3x 4e-h); b) (&)(e3%- 9e-% - 12x 8); C) j;(e2" - e-2U)g(x- u ) du; d ) J( sin u g(x - u ) du.

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(a)

Section 9.6, page 643 1. a) x = c1e5' cZe-*, y = 2cle5' - 3 ~ 2 e - ~ ' ; b ) cr(2cos 3t, cos 3t 3 sin3t) c2(2sin 3t, -3 cos 3t + sin 3t); C ) cl(2cos 3t, cos 3t 3 sin 3t) c2(2sin 3t, -3 cos 3t + sin 3t) + (t2,t2); d ) ~ ( 11,,1 ) c2eat(2cosbt, -cos bt 2b sin bt, -cos bt - 2b sin bt) + c3eat(2sin bt, -sin bt - 2b cos bt, 2b coo bt - sin bt), where o = b =

+

+

+

+ +

+ +

+

+

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&,

e) cle3'(1, 1) c2e2'(2, 1 ) (;)(-get - 4, -3et 2); f ) clet(l, -2, -1) cze-'(l,O, -1) c3e2'(4, -3, -1) ( f ) ( - 4 , 1 - 2t, 3 - 2t). 2. a) x = 2c1 cost 2c2 sin t 2c3eat 2 ~ 4 e - ~y' ,= cl cost c2 sin t - c3eatwhere a = b ) cleW3'+ c2et - t2. 3. a) x = e5' + eFst, y = 2e5' - 3e-5'; b ) ($)(18cos 3t - 2 sin 3t, 12cos 3t 26 sin3t); c) (t2,t2); d) ( I , 1, 1); e) e3'[c1(1- 2t, -2t) c2(2t, 1 2t)l; f ) et[cl(l+ t (t2/2),-t2/2, -t - 2t2)+ ~ 2 (t 3(t2/2),1 - 4t 3(t2/2),-13t 6t2) c3(t2/2,t - (t2/2),1 + 3t - 2t2)].

a;

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Section 9.7, page 646 1. a) X I = Al sin (2t a l ) ,x2 = 0 and xl = 0 , x2 = A2 sin (3t b) xl = -x2 = A1 sin ( t a l ) and xl = x2 = A2 sin ( f i t 4 2 ) .

+

+

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+42);

i

Answers to Problems Section 9.8, page 649 1. a) 1 2(x - 1) 3(x - 1)2

+ q ( x - 1)3+ ... ; 1 - $x2 + ;x4 b ) 3 + $ X 2 + $x3 - ;x4 + . . ; d ) x + x2 + &x4 + A x 5 . +

+

C)

7

m X

.

6.

2"x2" 2. a) c o [ l - 2 ~ ~22+ ~ ~ ~ + . . . + ( - 31. .). ("2 ~n -.l j + . . . I

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2n-Ix2n-I n-2) x4n + ' ' ' + (4n)(4n - 1)(4n - 4)(4n - 5 )

I

+.-.I;

x8 (4 (8 . 7)(4 . 3 x4n+l x9 +... + (4n + 1)(4n)(4n- 3)(4n - 4) ...( 5 4) 19 . 8)(5 . 4 )

b ) co[l +

- . 3) + 1 + + ~1 [ X + + +...I. 3. 1 - $ + 6- & + & + . . + cnxn + . ., where cn + (n + 2)cn+2+ ( n + 3 ) ( n + 2 ) ~ , , + ~ = 0 .4. y = l + $ + . . . . r = x + l ; + - a * . x5

3

X2n+~

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+...I.

6. c[x - %2 + 2 . 4 . 4 . 6 + ". + ( - 1 ) n 2 . 42 . 62 . . . (2n)2(2n+2) [This is 2c Jl(x),where J1 is the Bessel function of first kind of order 1.1 w (- 1)kx2k where m = fn, but m not a negative integer. " cxm Ck=O 4'k!(m + l ) ( m + 2 ) . .. (m + k ) ' this is J,(x), the Bessel function of first kind and [For rn a positive integer and c = order m .]

8. a)

where m = 0 (unless y is 0 or a negative integer) or m = 1 - y (unless y is a positive integer); the series converges for J x1 < 1; [For m = 0 the solution is cF(cr, B, y , x), where ~ t . F is the hypergeometric series.] L c b)

where rn = -a (unless cr - @ is a negative integer) or m = -B (unless B - a is a negative integer); the series converges for 1x1 > 1. h ( ~ ( a - 2 )... ( a - - 2 n + 2 ) ( ( r + l ) ... ( a + 2 n - 1 ) 9. a) co[l C g l ( - l ) x (2n)!

+

I

The series converge for Ix 1 < 1 or else reduce to polynomials. Section 9.9, page 656 1. a) 0 , t 3 / 3 ,(t7/63) (t3/3),(t15/59535) (2t11/2079) (t7/63) t3/3; b ) 1, 1 - (t2/2),g(t) = 1 log(1 - (t2/2)),1 - ~ ~ [ u / g ( u ) ] d u . 2. b ) x = cet2", c 2 0, x = ce-"J2, c < 0. 3. b ) x = (c - : x ) - ~and x = 0. 4. b ) x = f[c (2t/3)13J2and x = 0 , and combinations of the two forms of solution.

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729

Answers to Problems 'sra

Section 10.12, page 694

(-1)" 1. x2 sin t + E l 1 -ns sin nx e-"lr. 2. F o r w = f l , u = f i s i n x ( s i n t -?cost). Foro # f1, u = sinx(sinwt - wsint)/(l - w2).

Section 10.15, page 701 3. CZ1{ C z lcmnsin nx sin my e-c2(m2+n2)t

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+

1.

5. J,,,(&r) sin (a& r r)(clcos me c2 sinme), where cl and cz are constants. 6. CE, { CE'=o J ~ ( & r ) e-c2Amn'(amn cos me Pmn sin me)}. 8. cl coscrx + c2 sinax, 0 5 a < 00,A = aa.

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731

Index

Abel's formula, 461 (Rob. 10) Abel's summation of series, 611 (Rob. 5) Absolute convergence, 386,440,452,463,560 maximum, 146,172 minimum, 146,172 Absolute value of complex number, 438,532 Accelerating convergence of series, 410 (Prob. 7) Acceleration vector, 127 (Prob. 5) Addition of complex numbers, 532 of matrices, 19 of vectors, 46 A q ' s stress function, 615 Algebraic equation, 35,587 (Rob. 3), 640 Alternating series test, 391 Amplitude, 477 Analytic function, 430,444,544-624, 591-592,688 continuation, 550,565 along a curve, 547 at infinity, 572, 579 at a point, 547 $9 Angle, 50 Angular velocity vector, 186-187 (Rob. 16)

Annular domain, 567 Antiderivative. See Indefinite Integral Arbitrary constants, 626 Arc length, 122-123, 189, 195 (Probs. 6,7), 196-197 (Prob. lo), 203,242,247, 276278,543 Area element vector, 270,3 11-3 15,314 (Rob. 9) Area of parallelogram, 5, 12 plane, 216,228,278,286, 330 (Prob. 3) of surface, 243-249,305-306 of surface of revolution, 248 (Rob. 6) Argument of complex number, 532 (Fig. 8.1) principle, 583 Arithmetic mean, 217,230,475,558 Average value, 217,230,475,558

Base or basis vectors, 2,8,47,62,499 Bessel, differential equation, 514, 649-650 (Probs. 6,7), 701 (Rob. 4), 702 (Prob. 7) functions, 5 12-5 17,649-650 (Probs. 6.7). 697,701 (Prob. 4), 702 (Rob. 7) inequality, 477,500

734

Advanced Calculus, Fifth Edition Beta function, 459 Biharmonic equation and functions, 137, 143 (Robs. 4,5, 9), 614-616 Binomial coefficients, 444 series, 427-432 Binormal, 128 (Prob. 6) Boundary conditions, 603407,677 , , , point, 75.83 (Prob. 13) value problem, 603414,657-713 (Chap. 10) Bounded function, 78,146,153, 172,249,566 (Prob. 6), 571, 574 (Prob. 7) sequence, 168,378 set, 74, 168, 225 Bound vector, 176 Branch, 550-552

Comparison tests for integrals, 452 for series, 387-388, 397 (Prob. 13) Complementary function, 689 Complete ellipt~cintegral, 253 Complete orthogonal systems, 500,506 (Probs. 1,2), 507 (Probs. 11-15), , 511-515,517,681,696-700 Complex conjugate, 532 (Fig. 8.1) Complex number infinity, 572-574,579-581 Complex number system, 53 1 Component, 2,4,46,190, 197-200 Conditionally convergent, 387 Conformal mapping, 548,591-624 Connected open set, 75 Conservation of energy, 339,667 (Prob. 4), 67 1 (Prob. 2) of mass, 3 19 Calculus of variations, 705 Constant coefficients, 6 3 W 6 Cauchy criterion, 382,385,439,451 Continuity equation, 182, 319, 340,349-350 inequalities, 566 (Rob. 5) (Prob. 9) inequality, 53 (Prob. 7) Continuous function, 78, 81, 146, 153, 171, integral formula, 557-558 381,418,454,533,537 integral theorem, 553-556 to the left or right, 533 product of series, 407 Continuous spectrum, 700 residue theorem, 576,580 Contour lines. See Level curves Cauchy-KowaleskiTheorem, 703 Convergence in the mean, 501 Cauchy-Riemann equations, 347 (Prob. 2), of integrals, 218,249-253, 300,447456, 544-548,612,659 462465 Center of mass, 229,233, 313 (Prob. 3) interval, 423 Chain rules, 96-98, 101-104 of sequences, 167-171,377,410-417, Change of coordinates, 54-55 (Probs. 13, 14), 438 118, 187-197 of series, 375,383,410-417,440,560 of variables in integrals, 236-242, 331-338, Convex, 593,621 Convolution, 520,529 368 Characteristic equation, 35,640,664 Coordinates in space, 1 functions, 679,686,696,698,708 Coplanar vectors, 6 values, 35-38, 156,641,664,670,679,686, Cotangent space, 372 Coulomb's law, 343 696,699,700,704,707,708 Characteristics, 682 Cramer's Rule, 13 i Charge density, 182, 341 Critical damping, 668 (Prob. 9) Chebyshev polynomials, 507-508 ( h b . 15) Critical points, 145-161,601 ( h b .7), 707 Christoffel symbols, 205 Cross product, 5 Circle of convergence, 441,560,564 Cross ratio, 601 (Prob. 9) Circulation, 325,340 - 4 Curl of a vector field, 175, 182-186, 191-193, Closed curve, 270 if 207,212 (Prob. 18), 286,294, 300, region, 75 320-33 1,340-34 1,347 (Rob. 2). 367, set, 74, 83 (Robs. 12, 13), 160 (Rob. 13), 612 171-172 Curvature, 128 (Prob. 6), 213-214 Collinear, 6 (Probs. 24-25) Compact set, 172 Curve in space, 122-130

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Index Curvilinear coordinates, 119, 187-197,238, 246 Cylindrical coordinates, 120, 141, 195 (Prob. 6) Damped vibrations, 661,670 Definite integral, 215-219,253-257,272,304, 447,541-544,587 Degree, 100 (Prob. 1I), 330,443,574 (Prob. 4) of differential equation, 625 of differential form, 365 of mapping, 332-338 Deleted neighborhood, 569,572,574 (Probs. 7, 8) de l'H8pital's Rule, 130 (Prob. 19) Del symbol, 125, 179 Demoivre theorem, 550 Density, 230,233, 351,524 Dependent variables, 76, 106 Derivative, 83-167, 179-197,205-207,209, 221,230,254,367,421,454,506,534, 539,544-552,564 Determinants, 6, 9-17, 32, 94 Developable surface, 130-131 (Prob. 20) Diagonalization, 43,155-158,160 (Prob. 15) Difference, 437 (Prob. 9), 447 (Rob. 5), 671 (Prob. 5) equation, 671-672 (Probs. 5,6), 679, 704 Differentiable, 86 Differential, 86,539 area vector, 270, 311-3 12 equation, 137, 144 (Probs. 12, 13), 509, 514,625458 (Chap. 9), 657-713 (Chap. 10) forms, 212-213 (Probs. 23,24), 364-374, 446 Differentiation of integrals, 253-257,454 of series, 421,426-427,506,561 Diffusion of heat, 677 Dimension of vector space, 62,67 Dirac delta function, 524-525 Direction angles, 4 cosines, 4 Directional derivative, 131-135,149-150, 179,539 Dirichlet problem, 603412,700 Distance, 2, 3,9,46,81, 122, 189,202. See also Arc length

Distributive law, 4,5,20,24,47,314 (Prob. 9), 498 Divergence, of sequences, series, integrals. See Convergence Theorem, 286,3 14-3 19 of a vector field, 175, 181-197,206,286, 300,314-319,348-350 (Robs. 7,8,9), 367,612 Diverges to infinity, 379 Domain, 75,538 Dot product, 3,46 Double Fourier series, 517-518.699 Double integrals, 145 (Rob. 14). 225,231, 238,262,337 (Prob. 4) Double layer, 352, 359, 361 Doubly connected, 293,556 Dynamics, 339,644-646 Eigenvalues. See Characteristicequation, values Eigenvector, 35-39,61 (Prob. 12), 156 Elasticity, 137 Electric force, 178, 182,340-343 Electromagnetic waves, 342,677 Electromagnetism, 182,340-343 Electrostatic potential, 186 (Prob. 8), 342, 347 (Prob. 3), 610 (Prob. 2) Elementary analytic functions, 532 Ellipsoid, of inertia, 235 (Prob. 8) Elliptic differential equation, 676 functions, 224-225 (Probs. 5-7) integrals, 222-223,253 Empty set, 83 (Prob. 13) Energy, 74, 339,671 (Prob. 2) Entire function, 566 (Prob. 6) Entropy, 346 Equation of state, 344 Equilibrium, 610 (Prob. 3), 658,661,666, 670,677,688489,699,704 Equipotential lines, 613 Error, estimation of, 220 (Probs. 7,8), 230, 399405,433 function, 225 (Prob. 8), 252 (Prob. 1) Essential singularity, 571,573,574 (Prob. 8), 579 Euclidean, n-dimensional space, 46,52 vector space, 52,6743,497,672473 (Prob. 8) Euler identity, 442,532 Euler-Mascheroni constant, 459

735

736

Advanced Calculus, Fifth Edition Euler's theorem on homogenebus functions, 100 (Prob. 11) Even function, 480 Exact differential equation, 629 (Prob. 8) Existence theorems for differential equations, 627428,631,637,651457 Exponential decay, 661,665,670,676, 685, 689 Exponential function, 432,532,547-552, 597-598 Extended ratio test and root test, 397-399 Extended z plane, 573,580 Exterior (wedge) product, 366 derivative, 367

-

Factorial, 458-459 Finite element method, 706 z plane, 573 First and second fundamental forms, 212-213 (Probs. 23,24) First difference, 437 (Prob. 9). 671 (Prob. 5) law of thermodynamics, 344-345 Fluid motion. See Hydrodynamics Flux, 315, 318 Force, 268, 350,361 Forced motion, 662,671,677,690 Fourier cosine series, 480 integral and transform, 5 19-523,700 series, 467-530 (Chap. 7), 565,611 (Probs. 4, 5) sine serles, 480,672-673 (Prob. 8), 679,693 Fourier-Bessel series, 5 13 Fourier-Legendre series, 5 10 Free energy, 348 (Prob. 6) Frenet formulas, 128 (Prob. 6) Frequency, 68 1,700. See also Resonant frequencies Function, 73,80-83,91, 171-172,233,236 (Prob. 1l), 532, 591 of complex variable, 531-624 composite, 96, 101 of several variables, 73 Functional, 528, 706 dependence, 161-167 equation, of the Gamma function, 458 equation, permanence of, 547 Fundamental lemma, on differential, 87 mode, 68 1 theorem of algebra, 615 (Prob. 3) theorem of calculus, 216, 221

GaddNriiori, 458-459 Gas, theory of, 117 (Prob. 8), 344-347 Gaussian elimination, 32-35,6347 Gauss-Jordan method, 32 Gauss's theorem, 286, 3 15 General exponential function, 550-551 General solution, 626 Generalized functions (distributions), 523-529 Geometric series, 389, 561 Gradient, 125, 179, 185-186,206,293-294, 326 Gram-Schmidt orthogonalization process, 50, 51 1 Gravitation, 177, 180 (Prob. 4), 286 (Probs. 3,4), 361 Greatest lower bound, 169, 172 Green's function, 360-362,708,710 (Prob. 9) identities, 302-303 (Probs. 11, 12), 320-321 (Prob. 4), 358-362 theorem, 283, 297-298 Grouping of series, 405 Harmonic function, 137, 185, 186 (Prob. 8), 206,320 (Prob. 4), 340,343,347 (Prob. 2), 353-364,486 (Prob. 3), 565,603412,659,676 motion, 468,64446,661,664,670,672 (Prob. 7), 676 series, 389 series of order p, 389 Heat conduction, 343-344,665,677,685689 Heat equation. See Heat conduction Heaviside unit function, 524 Helmholtz theorem, 362 Hennite polynomials, 5 15 Hessian matrix, 180 (Prob. 9) Higher derivatives, 135,138, 142 Homogeneous, differential equation, 629 (Prob. 7), 630, 638 equations, 14,33 functions, 101 (Rob. 1l), 330 polynomials, 443 problem, 677,689 Hydrodynamics, 100 (Prob. 12), 175, 181, 183, 185 (Prob. 2), 257 (Prob. 7), 315, 319,325,347-350 (Probs. 2,7,8,9), 617 (Probs. 1,2), 677 Hyperbolic equation, 676 functions, 532,548-549

Index Hypergeometric equation, 650 (hob. 8) Hyperplane, 62 Identities of Green. See Green's function, identities Imaginary part, 439,531 Implicit functions, 105-1 17 Improper integrals, 218, 249-253, 300, 447-456,462-465 Incompressible flow, 176, 182, 185 (Prob. 2), 318,340,347 (Prob. 2), 612414,617 (Probs. 1,2) Indefinite integral, 216,221, 256 (Prob. 5), 535,555 Independence of path, 287,325-327,553-555 Inequality, 49,216,230,476,498,500,566 (Prob. 5) Infinite sequences, 376-385,410417, 418-422,438,442 Infinite series, 375-465 (Chap. 6). See also Fourier series, Power series Infinity, complex, 572-574.579-582 Inflection point, 145 Initial conhtions, 627,637,680,685 value problem, 627,703,710 (Prob. 5) Inner product, 3,46,497,517,672 (Prob. 8) Input, 662, 667 Insulated boundary, 678 Integral, complex, 535,541-544,553-561, 575-583,587-590 depending on a parameter, 253 function, 566 (Prob. 6) test, 388 of vector function, 233,236 (Prob. 11) Integration by parts, 217 Integration of series, 420,505,561 Intermediate value theorem, 82 (Prob. 8), 173 (Prob. 5) Internal energy, 344 Inverse functions, 118-122,435,549-552, 583,592-593 points, 602 (Prob. 10) trigonometric functions, 551 Irrational numbers, 167 Irrotational, 183,327-328,340,347 (Prob. 2). 612 Isolated boundary point, 569 singularity, 569,572, 579 Isolation of zeros, 572 Iterated integral, 227,261

Jacobian determinant, 94, 119,237,348-349 (Prob. 7), 660 matrix, 91, 120 Jacobi polynomials, 515 Jump discontinuities, 47 1

Kinetic energy, 281, 339,671 (Prob. 2) Kronecker delta, 18 Kronecker integral, 338 (Prob. 6) Lagrange multipliers, 154,707 Laguerre polynomials, 5 16 Laplace equation. See Harmonic function Laplace transform, 457,521,528,635 (Prob. 16), 643 (Prob. 3) Laplacian, 136, 140-142, 184-185, 192,206, 675,704. See also Hannonic function Laurent series, 566-581 Least squares, 160 (Prob. 1I), 476,501 Least upper bound, 169, 172.413 Legendre differential equation, 509,650 (Prob. 9), 697 polynomials, 508-512,516 (Probs. 1-9), 650 (Prob. 9), 697,702 (Prob. 7) Leibnitz Rule for derivatives, 508 for integrals, 254,256 (Prob. 5), 349 (Prob. 8) Length of a curve. See Arc length Level curves, 76, 161,593-600 surfaces, 77 Liebrnann's method, 709 (Prob. 2) Limit, 78, 167,533,538 in the mean, 501 Line, 9 Line integral, 267-305,321-331,339-347, 541-544,553-561,575-583 Linear differential equations, ordinary, 629 (Prob. 9). 629-646 with constant coefficients, 630-646 ,j partial, 657-713 (Chap. 10) Linear equations, 8, 13-15,19,32-35,55, 63-66 function, 372 mapping, 55,91 operators, 183-1 84,630 Linear, fractional transformation, 595 integral transformation, 595 Linearization, 106 Linearly dependent, 6,47,498,631,638,641

737

738

Advanced Calculus, Fifth Edition Liouville, substitution, 144 (Prob. 13) theorem, 566 (Prob. 6) Lipschitz condition, 652 Logarithmic derivative, 582 function, 549-552 potential, 252 (Prob. 2), 350,699 residue, 582-584 Lorentz condition, 341 Lower limit of sequence, 379-381

syrnmetric.40 ' triangular, 32 zero, 18 Matrix function, 638-644 Maxima and minima, 145-160,446,705, 707 Maxwell equations, 341 Mean value. See Average value Mean value theorem, 217,230,233,445 Mechanical system of N particles, 662,668 Mesh, 215, 225 Metric space, 81 Minkowski inequality, 498 Moment of inertia, 229,3 13 (Prob. 3) Monotone sequence, 167,378 Morera's theorem, 554 Multilinear, 372 Multiple integrals, 232 M-test for integrals, 454 for series, 416 Multiplication of matrices, 21 of series, 407-409 of vector by scalar, 2,46 of vectors, 3-5,46 Multiplicity of zero, 571,573,577, 581,582 Multiply connected, 292,297-300,555, 558

Maclaurin series, 428,561-565 Magnitude of a vector, 2, 3,46,211 (Prob. 15) Mapping. See Conformal mapping; Transformation of coordinates; Function; Linear equations, mapping Mass, 229,233, 313 (Prob. 3), 349 Matrices, 18-71,436 addition, 19 augmented, 32 characteristic equation, 35 column vectors, 19 determinant of, 26 diagonal, 36 eigenvalues, 35-38, 156,641 eigenvectors, 35-38,61 (Prob. 12), 156, 64 1 entry, 18 identity, 18 Nabla symbol, 125, 179 inverse of, 26,33 Negative direction, 273-275 Jacobian, 91,207 triple, 5 kernel of, 58,64 Neighborhood, 74-76,538 of maximum rank, 65 Newtonian potential, 253 (Rob. 3), 361 minor of, 65 Newton's Second Law, 281 multiplication, 21 Nonhomogeneous differential equations, nonsingular, 28 63 1-646 nullity, 64 Nonorientable surface, 307 order of, 18 Norm, 2-3,46,211 (Prob. 15), 497,517, orthogonal, 41 672-673 (Prob. 8) orthogonally congruent, 42 Normal component of acceleration, 128 permutation, 31 (Prob. 11) (Prob. 7) powers of, 24,28 to curves, surfaces, 8, 125, 128 (Probs. 6,7), product of, 21 245 rank of, 33,6347 Normalized orthogonal systems, 497 row echelon form, 33 Normal modes, 645,6644565,670,677479, row vectors, 18 686,700, scalar times, 19 nth term test, 386 ' similar, 36 Numerical evaluation of integrals, 218-219, singular, 28, 29 221-223,279 (Prob. 6) square, 18 methods for differential equations, 649, subtraction. 20 703

Index Odd function, 480 One-to-one transformation,58, 118-122,237, 240,331-336,591-596,617-623 Open interval, 76 region, 75 sets, 74 Operational method, 634 (Prob. 13) Order of differential equation, 625,676 of pole, 570-573,576,582 Ordinary differential equauons, 625458 (Chapter 9) existence of solutions, 627428,631,637, 65 1-657 Orientable surface, 306-307 Orientation in space, 5, 364 Orthogonal curvilinear coordinates, 189, 196 (Prob. 10) basis, 50 functions, 495,499,681,696 vectors, 3,42,50,672-673 (Prob. 8) Orthonormal, 50,497499 Osculating plane, 127 (Prob. 5 ) Output. See Input Overtone, 468,681

integral formula, 257 (Prob. 8), 361,362, 603,607,611412(Probs. 5,6) summation of series, 611 (Prob. 5 ) Polar coordinates, 1 19, 122 (Prob. 8) moment of inertia, 229 Pole, 570-587 Polynomial, 80, 538,547 Position vector, 2 Positive definite, 45 (Prob. 13), 160 (Probs. ..% 13-15), 196 (Prob. 10) Positive direction, 273-275 triple of vectors, 5 Potential, 340-343 Potential energy, 339,670,705 theory, 350-364.700 Power function, 55 1 series, 422-437,441,560-574 Primitive. See Indefinite integral Principal, normal, 128 (Prob. 6) part, 570,574 (Probs. 2,6), 585 value of function, 550-552 value of integral, 455-456,463 Products of inertia, 235 (Prob. 8) Properly divergent, 384

Parabolic equation, 676 Parallel, 5 axis theorem, 235 (Prob. 7) Parametric equations, 9, 122-123, 129 (Prob. 16), 243 Parseval's equation, 494 (hob. 2), 500, 505 Partial derivatives, 83, 135 Partial differential equations, 137, 626, 659-712 (Chap. lo), Partial fractions, 584-586 Partial sums of series, 383 Particular solution, 626,63 1 Periodic functions, 467 Permanence of functional equation, 547 Permutations, even or odd, 11, 17 (Prob. 16) Perpendicular, 3,50 Phase angle, 477 Piecewise smooth, 272,307,542 Planar sound waves, 676 Plane, 8 area. See Area of parallelogram, plane Point discontinuity, 250 Poisson equation, 320 (Prob. 4), 342,358, - .... 362,616,677,708

Quadratic form, 40, 154-156,202,644,670

Radius of convergence,423,44 1,560-565 of curvature, 128 (Prob. 6) Range, 58,171 Rational functions, 80,532,547,570,574 (Prob. 41,584-586,594-596 numbers, 167 Ratio test, 390,397,460 (Prob. 3) Rayleigh-Ritz method, 706 Real number system, 167 Real part, 531 Rearrangement of series, 406 -. Reciprocal transformation, 595 Recursion formula, 509,5 14,648 Region, 75 Relaxation method, 709 (Prob. 2) Removable singularity, 569,572,574 (Probs. 7, 8), 585 Residues, 575-583,587 (Probs. 3-8). 587-590 Resonance, 668 (Prob. 9) Resonant frequencies, 645,664,670,679, L 686,696,700,704 Riemann integral, 2 15,258-260 'A$

740

Advanced Calculus, Fifth Edition Riemann theorem, 574 (Prob. 7) Right-handed systems, 5 Rodrigues, formula of, 508 Root test, 392,398,460 (Prob. 5) Roots, 551 Rotation-stretching,594 Row space, 63 Scalar, 2 field, 179 product. See Dot product; Inner product times vector, 2,46 triple product, 11 Schwarz inequality, 49,498 Schwarz-Christoffeltransformation, 619-623 Second derivative, 135, 138, 142 Second difference, 437 (Prob. 9), 447 (Prob. 5), 671 (Prob. 5) Second directional derivative, 149 Second law of thermodynamics, 347 Self-adjoint, 701 (Prob. 2) Sense-preserving,591-592 Separation of variables, 629 (Prob. 6), 699 Sequences. See Infinite sequences Series. See Infinite series Set, 74 Side condition, 154,707 Simple closed curve, 270 harmonic motion, 468,645,661 Simply connected, 292, 327,550, 553 Simultaneous differential equations, 636-646 Simultaneous equations, 13-15, 32-35,55,67 (Robs. 6-9), 105-1 17 Single layer, 352, 358,361 Singular case, 697 point, 569, 579,648 Slit domain, 614 Small vibrations, 644 Smooth curve, 270 function, 472 surface, 305 Solenoidal, 184,328,331 (hob. 6), 340,660 Solid angle, 337 (Prob. 5) Sound, 468,676 Specific heat, 344,665 Spectrum, 679,700 Spherical coordinates, 120,194-197 Standard basis, 8,49 Stationary flow, 175,612-614. See also Hydrodynamics

Step-by-step integration of differential equation, 709 (Prob. 1) Stereographicprojection, 573 Stirling approximation to factorial, 459 Stokes's derivative, 101 (Prob. 12), 185 (Prob. 2). 257 (Prob. 7), 340 theorem, 285,323,369 Straight line, 9, 124 Stream function and lines, 347 (hob. 2), 613 Stress function and tensor, 615 Sturm-Liouvilleboundary value problems, 144 (Prob. 13), 696-4598 Submersion, 166 Subsequence, 169 Subspace, 62 Substitution in differential equations, 144 (Probs. 12, 13) Superposition,493,695 (Prob. 6), 708 Surface, 124, 129-130 (Probs. 16, 20), 305-306 area, 243-249,305-306 area element, 270,311-3 12,314 (Prob. 9) integrals, 308-3 12 System of equations. See Simultaneous equations Table of integrals, 222 Tangent line, 122-133 plane, 125-126 space, 371 vector, 122-124, 127-130 (Probs. 5.6, 19), 592,602 (Prob. 14) Tangential developable, 130 (Prob. 20) Taylor formula with remainder, 154,436 (Probs. 3,4), 445 Taylor series, 428,444,544,560-565,646 Temperature, 343-347,610-611 (Prob. 3), 6 17 (Prob. 5.). See also Heat conduction: Thermodynamics Tension, 675 Tensors, 197-2 14,370-374 algebra of, 203 alternating, 209-213, 370 contraction of, 204 contravariant, 198-199 covariant, 198-199 derivatives, 205 exterior product (wedge product), 210,371 invariant, 201,371 metric, 203

r

Index on a surface, 208-209 order, 201 product, 204 Thermodynamics, 117 (Rob. 8), 344-347 Torsion, 128 (Prob. 6) Total differential, 87 Transformation of coordinates, 54-55 (Probs. 13, 14), 118-120, 187-197, 236,331,591-624 Translations, 55 (Rob. 14) Trapezoidal rule, 2 19 Triangle inequality, 49, 8 1, 531 Trigonometric, functions, 532,547-552, 564 series, 467,611 (Probs. 4.5). See also Fourier series Triple integrals, 232,240 Triply connected, 293,556 orthogonal family of surfaces, 189 Trivial solution, 14-15 Umbilical point, 214 (Rob. 25) Unbounded regions, 252,700 Undetermined coefficients, 635 (Prob. 14) Uniform continuity, 258 convergence,41 1-417,453455,489-493, 560,567 Uniqueness property, 486,501,s 10 Unit vector, 2,46,499 Upper limit of sequence, 379-381 Variable coefficients, 647,695 Variational method, 705-707 Variation of parameters, 632, 639-640, 691

Vector, field, 176,612 functions, 81,91, 233, 236 (Prob. 1l), 534 identities, 5, 11, 17 (Prob. 12), 183-186 product, 5,212 (Prob. 19) space, 52,67,497,672-673 (Prob. 8) triple product, 5 Vectors, 1-1 1, 18-19,4655, 175-214 (Chap. 3), 267-374 (Chap. 5) Velocity potential, 340,347 (Rob. 2), 612 vector, 122, 128 (Probs. 6-7) Vibrations of a circular membrane, 702 (Prob. 5) of a spring, 661-674 of a string, 468,486 (Prob. 2), 674,678-685 Volume, 11,51, 104,227,228,233 Wave equation, 342,677,678,698 length, 683 velocity, 683 Weierstrass-Bolzanotheorem, 169 Weierstrass and Casorati, theorem of, 574 (Prob. 8) Weierstrass M-test, 416,454 Weight function, 507 (Rob. 15), 513 Winding number, 301 (Prob. 5), 583 Work, 268,281,305,339,346 Wronskian determinant, 632,636 (Prob. 17) Zero mapping, 59 Zero matrix, 18,59 Zeros of function, 571-573,574 (Prob. 4), 577,581-584,587 (Probs. 3,6,7) Zero vector, 3, 19 Zeta-function of Riemann, 389

741

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